
Class g^% S 
Book '':JXJ3 



INTRODUCTION 

« TO 

MENSURATION :^ „ 

PRACTICAL GEOMETRY. 

BT 

JOHN BONNYCASTLE, 

OF THB BOTAL MILITARY ACASEMT, WOOtWICH, 



TO WHICH ARE ADDED, 

A TREATISE ON GUAGING: 

AJTD ALSO THE -a^ 

MOST IMPORTANT PROBLEMS IN MECHANICS. 



BY JAMES RYAN, 

Autbor of a Treatise on Alj^bra, the New American Grammar of AstroBoasy, 
The Differential and Integral Calculus, &c 




KIMBER & SHARPLESS, No. 8 SOUTH FOURTH ST. 



8TEHEOTTPED BT J. HOWE. 



1834. 



Entered according to the Act of Congress, in the year 1833, by 
KiMBER & Sharpless, in the Clerk's Office of the District Court of the 

Eastern District of Pennsylvania. 









/ 



PREFACE 

TO THE LONDON EDITION. 



The art of measuring, like all other useful inven- 
tions, appears to have been the offspring of want and 
necessity ; and to have had its origin in those remote 
ages of antiquity, which are far beyond the reach of 
credible and authentic history. Egypt, the fruitful 
mother of almost all the liberal sciences, is imagined 
likewise to have given birth to Geometry or Mensura- 
tion; it being to the inundations of the Nile that we 
are said to be indebted for this most perfect and de- 
lightful branch of human knowledge. 

After the overflowings of the river had deluged the 
country, and all artificial boundaries and land-marks 
were destroyed, there could have been no other method 
of ascertaining individual property, than by a previous 
knowledge of its figure and dimensions. From this cir- 
cumstance, it appears highiv probable, that Geometry 
was first known and cultivated bv the ancient Egyp- 
tians; as being the only science which could administer 
to their wants, and furnish them, with the assistance 
-hey required. The name itself signifies properly the 
art of ?neasuring the earth: which serves still further to 
confirm this opinion, especially as it is well known that 
many of the ancient mathematicians applied their 
geometrical knowledge entirely to that purpose, and 
that even the Elements of Euclid, as they now stand, 
are only the theory from whence we obtain the rules 
and precepts of our present more mechanical practice. 

But to trace the sciences to their first rude beginnings, 
s a matter of learned curiosity, which could afford 
ut little gratification to readers in general. It is of 
iuch more consequence to the rising generation to be 
jformed that, in their present improved etaie, they ar^ 



exceedingly useful and important. And in this respect, 
the art I have undertaken to elucidate is inferior to 
none, arithmetic only excepted. Its use in most of the 
different branches of the Mathematics is so general 
and extensive, that it may justly be considered as the 
mother and mistress of all the rest, and the source from 
whence were derived the various properties and prin- 
ciples to which they ow-e their existence. 

As a testimony of this superior excellence, I need 
only mention a few of those who have studied and im- 
proved it ; in which illustrious catalogue we have the 
names of Euclid, Archimedes, Thales, Anaxagoras, 
Pythagoras, Plato, Apollonius, Philo, and Ptolemy, 
amongst the ancients : and Huygens, Wallis, Gregory, 
Halley, the Bernouillies, Euler, Liebnitz and Newton, 
amongst the moderns ; all of whom applied themselves 
to particular parts of it, and greatly enlarged and im- 
proved the subject. To the latter especially we are 
indebted for many valuable discoveries in the .higher 
branches of the art ; which have not only enhanced 
its dignity and importance, but rendered the practical 
application of it more general and extensive. 

The degree of estimation in which the art was held 
by these, and other eminent characters, will, in gene- 
ral, it is apprehended, be thought a sufficient encomium 
on its merits. But, for the sake of young people, and 
those of a confined education, it may not be amiss to 
give a few more instances of its advantage, and show 
that its importance in trade and business is not inferior 
to its dignity as a science. Artificers of almost all de- 
nominations are indebted to this invention for the estab- 
lishment of their several occupations, and the perfec- 
tion and value of their workmanship. Without its 
assistance, all the great and noble works of Art would 
have been imperfect and useless. By this means the 
architect lays down his plan, and erects his edifice ; 
bridges are built over large rivers; ships are con- 



atructed; and property of all kin^s is accurately mea- 
sured, and justly estimated. In short, most of the 
elegances and conveniences of life owe their existence 
to this art, and will be multiplied in proportion as it is 
well understood, and properly practised. 

From this view of the subject, it is hardly to be ac- 
counted for, that, in a commercial nation, like our own, 
an art of such general application should have been so 
greatly neglected^ Mechanics of all kinds, it is well 
known, are but ill acquainted with its principles ; and 
those who have been the best qualified to atibrd them 
any assistance, have thought it beneath their attention. 
Till within a few years past, there could not be found 
a regular treatise upon this subject in the English 
language. Some particular branches, it is true, had 
been greatly cultivated and improved ; but these were 
only to be found in their miscellaneous state, inter- 
spersed through a number of Large volumes, in the 
possession of but a few, and in a form and hmguage 
totally unintelligible to those for whom they were 
more immediately necessary, 

Dr, Hutton was the first person, in this country, who 
undertook to collect these scattered fragments, and to 
treat of ihfi subject in a scientific, methodical manner. 
A small treatise by Hawney, and some others of little 
note, had indeed been long in the bands of the public; 
but these were extremely defective, both in matter and 
method; neither the principles nor practice of the art 
being, properly or c: early explained. Before the publi- 
cation of the treatise above mentioned, Mr. Robert- 
eon's may be considered as the only book, of any value, 
that could be consulted, either by the arti<an or mathe- 
matician; and had he given the theory as well as the 
practice of the art, and divested his rules and exam- 
ples of their algebraical form, there would have been 
no want of any other elementary treatise. 

To these two writers I am greatly indebted for manj 



things in the following pages, and am ready to ackn<vw- 
ledge, that I have used an unreserved freedom in se- 
lecting from their works, wherever 1 found them to an- 
swer my purpose. To Dr. Hutton I am particularly 
obliged, and am so far from desiring to supersede the 
use of his performance by this pubhcation, that I only 
wish it to be thought a useful introduction to it. His 
treatise is excellent in its kind ; and had it been as 
well calculated for the use of the uninformed Artist as 
it is for the Mathematician, the following compendium 
had certainly never been published. 

The method I have observed in composing this work, 
is that which was used in the " Scholar^s Guide to Arith- 
metic ;" and, as my object has been to facilitate the 
acquirement of the same kind of useful knowledge, I 
am not without hopes of its being received with equal 
candor and approbation. 

In school-books, and those designed for the use of 
learners, it has always appeared to me, that plain and 
concise rules, with proper exercises, are entirely suffi- 
cient for the purpose. In science, as well as in morals, 
example will ever enforce and illustrate precept; for 
this reason, an operation, wrought out at length, will 
be ^ound of more service to beginners than all the te- 
dious directions and observations that can possibly be 
given them. From constant experience I have been 
confirmed in this idea, and it is in pursuance of it that 
I have formed the plan of this publication. I have not 
been ambitious of adding much new matter to the sub- 
ject : but only io arrange and methodize it in a man- 
ner more easy and rational than had been done before. 

The text part of the work contains the rules in words 
at length, with examples to exercise them ; and, jr or- 
der that the learner may not be perplexed and inter- 
rupted in his progress, the remarks and demonstrations 
are confined to the notes, and may be consulted or not> 
as shall be thought necessary. To those who would 



wish not to take things upon trust, but to be acquainted 
with the grounds and rationale of the operations they 
perform, they will be found extremely serviceable: 
and for this purpose I have endeavored to make them 
as easy as the nature of the subject would admit. But 
they can be consulted only by such as have made a 
previous acquaintance with several other branches of 
mathematical learning. 

Some of the most difficult rules relating to the sur- 
faces of solids, &c. could not be conveniently given, 
but by means of algebraical theorems ; and as this was 
foreign to my purpose, I have not scrupled to omit 
them ; being well persuaded that what is done upon 
that head will be fully sufficient to answer most practi- 
cal purposes. In the Practical Geometry, likewise, 
which is prefixed to this treatise, such problems only 
are introduced as were known to be most intimately 
connected with the subject. And as this part of the 
work is a proper and necessary introduction to the rest, 
I have spared no pains in making it as clear and intel- 
ligible as possible. 

Upon the whole, I have endeavored to consult the 
wants of the learner, more than those of the man of 
science. And if I have succeeded in this respect, my 
purpose is answered. I have not sought for reputation 
as a mathematician, but only to be useful as a tutor. 

N. B. The favorable reception this work has met 
with, has induced me in this edition to make such 
alterations and additions as have since occurred to me, 
and which are such as I hope will render it still more 
acceptable to the public. 

Royal Academy, Woolwich, 
July 14, 1807. 



ADVERTISEMENT. 

The favonrable reception and ^eat demand for Bonnycas- 
tle's Mensuration and Practical Geometry, since its first 
publication in this country, induced me to publish the present 
edition, which contains not only the whole of that valuable 
work, but all that is most useful in Hutton, Hawney, Ingram, 
and other modern works on the same subject. 

To this edition is also added an article on Mechanics and 
Dynamics, containing the principal problems in Brunton's Me- 
chanics : — that is, Fulling Bodies ,- the Pendulum ,- the Lever ^ 
th% Wiieel and Axle^ the Pulley, tJie Inclined Plane, the Wedge, 
and the Screw, which are usually called the six Mechanical 
Powers ; Velocity of Wheels ,- Steam Engine ,- Water Wheels y 
and Pumps, 

JAMES RYAN. 

New York, Oct, 1st, 1833. 



CONTENTS. 



PRACTICAL GEOMETRY. 

Production, Page I 

Definitions, 13 

Geometrical Problems, 21 

OF THE MENSURATION OF SUPERFICIES. 

To find the area of a parallelogram, 51 

To find the area of a triangle, 54 

To find the sides of a right-angled triangle, 57 

To find the area of a trapezium, 59 

To find the area of a trapezoid, 60 

To find the area of a regular polygon, 61 

To find the diameter and circumference of a circle, . 64 

To find the length of any arc of a circle, 67 

To find the area of a circle, 70 

To find the area of a sector of a circle, 73 

To find the area of a segment of a circle, 76 

To find the area of a circular zone, 80 

To find the area of a circular ring, 82 

To find the areas of limes, 83 

To find the area of an irregular polygon, 84 

Promiscuous questions cf lines and areas, 86 

OF THE CONIC SECTIONS. 

Definitions, 90 

To describe an ellipsis, 93 

Cf an ellipsis and its parts, 94 

To describe a parabola, 102 

Of a parabola and its parts, 103 

To describe an hyperbola, 107 

Cf an hyperbola and its parts, ......•..•.....•.• 108 



OF THE MENSURATION OF SOLIDS. 

Pagk 

Definitionsy 118 

To find the solidity of a cube., 121 

To find the solidity of a parallelopipedon, 122 

To find the solidity of a prism^ 123 

Of a cylinder and its parts, 124 

Of the cone and pyramid and their partSy 128 

To find the solidity of a wedge, 138 

To find the solidity of a prismoid, 140 

Of the sphere and its parts, 142 

Of a spheroid and its parts, 148 

Cf a parabolic conoid and its parts, 155 

Of an hyperboloid and its parts, 157 

Miscellaneous questions in solids, 161 

Of the regular bodies, 165 

Of cylindric rings, 174 

Of Artificers* work, 177 

Cf Bricklayers' work, 178 

Of Masons' work, 181 

Of Carpenters' work, 183 

Of Slaters' and Tilers' work, 186 

Of Plasterers' work, 188 

Of Painters' and Glaziers' work, 190 

Of Paviors' work, 192 

Of vaults, domes, <^c 1 94 

Of the Carpenter's rule, 203 

Of Board and Timber measure, 207 

Specific gravity, 216 

Miscellaneous questions, 222 

A table of the segments of a circle, 231 



appendix. 

Page 

Of Gauging, .237 

Of the Gauging Rule, ib- 

Of the use of the Gauging Rule, 239 

Of the Gauging or Diagonal Rod, 241 

Of Casks, as divided into varieties, 242 

To find the content of a cask of the first form,, , . ib. 

the second form, . 244 

To find the content of a cask of the third form,, , 245 

the fourth form, . 246 

To find the content cf a cask by four dimensions, . . 247 

To find the content of any cask from three dimensionSi 248 

OF THE ULLAGING OF CASKS. 

To find the ullage by the Sliding Rule, 250 

To ullage a standing cask by the pen, ib. 

To ullage a lying cask by the pen, 251 

Of Guaging Casks by their mean diameters, 252 

To find the Tonnage of a Ship, 255 

Of Falling Bodies, 256 

Of the Pendulum, 258 

Of the Mechanic Powers, 8fc 259 

Velocity of Wheels, 266 

Of the Steam Engine, 268 

Of Water Wheels, 276 

Of Pumps, 284 



TABLES 



DIFFERENT MEASURES USED IN THIS WORK. 



Lineal Measures. 
12 inches make 1 foot. 

3 feet 1 yard. 

6 feet 1 fathom. 

16J feet, or ? S 1 pole, 

5| yards, S r o'' ^°^- 
40 poles 1 furlong. 

8 furlongs ... 1 mile. 



Square Measures. 
144 inches make 1 foot. 

9 feet 1 yard. 

36 feet 1 fathom. 

272^ feet ) U pole 
or 30;^ yds. \ ) or rod. 

1600 poles 1 furlong. 

64 furlongs . . 1 mile. 



JSfote ^The chain made use of in measuring land, com- 
monly called Gunter's chain, is 4 poles, or 22 yards m 
length, and consists of 100 equal links, each link bemg A\ 
of a yard=.66 of a foot, or 7.92 inches long. 

An acre of land is also equal to 10 square chains ; that 
is, 10 chains in length, and 1 in breadth ; or it is 4840 
square yards, or 160 square poles, or 100,000 square links. 



Note also that in Land Measure, 
40 perches, or ) ^^^^ j ^^^^ 
square poles ^ 
4 roods 1 acre. 



And in Cubic Measure, 
1728 inches make 1 foot 

27 feet 1 yard. 

166§ yards 1 pole. 



Other Measures. 
, inches make 1 !?allon ale measure. 



282 cubic i 

QOl A gl""" vviwv. ..^v..... 

Tit^l 1 gallon dry measure. 

I28^cubic feet, or 8 feet in length and i 

4 in breadth and 4 m height K 

34| cubic feet, or 16^ feet in length, j 

li in breadth, and 1 in height \ 



, 1 gallon wine measure. 



, 1 cord of wood- 
, 1 perch of stone. 



A TREATISE 



THE ART OF MEASURING^. 



INTRODUCTION. 



DECIMALS. 

If the numerator and denominator of a fraction be multi- 
plied or divided by any number, its value will not be alter- 
ed; thus, ^=tV' ¥=t¥o-' iFo=TTrf 0' ant^ so on. Hence, 
it is evident, that we can reduce a fraction to another equi- 
valent one, having a given denominator. It likewise fol- 
lows, that a fraction may be reduced to another equivalent 
one, whose denominator shall be 10, or some number pro- 
duced by the continued multiplication of 10, by annexing 
ciphers to the numerator and denominator, and dividing 
both, (with the ciphers annexed,) by the original denomi- 
nator. 

Thus, the fraction i=-j^^ and dividing both the numera- 
tor and denominator of the fraction ^^ by 5, the original 
" ^nominator, we should have yf =y^o'* 

Again, the fraction i=}^^ ; and dividing both the nu- 
merator and denominator by 4, we shall have iff =tVo' 
Also, 3 — 30^^. . gjjj dividing both the numerator and deno- 
minator of the fraction ffff , by 8, the original denomina- 
tor, we shall have im=iWo 5 hencef =y3j^^o- ? ^^^ so on. 

Fractions whose denominators are 10, 100, 1000, &,c. 
A 



are called decimal fractions ; and when these fractions arc 
written without the denominator, they are usually called 
decimals ; and to denote the value of a decimal, a point is 
prefixed to as many figures of the numerator as there are 
ciphers in the denominator. 

Thus, the decimal fraction -^-^ is written .2, the decimal 
fraction y2_5_ jg written .25, the decimal fraction y^/^ is 
written .375, and so on. The expressions .2, .25, .375, 
&c. are called decimals. Hence, it is evident, that the 
figures next the decimal point indicates tenths, the next 
figure hundredths, the next thousandths, and so on. 

The decimal .2 is read two-tenths ; the decimal .25 is 
read twenty-jice hundredths ; the decimal .875 is read eight 
hundred and seventy-jive thousandths ; and so on. 

Since, y2_=yVF=i¥o'o = rV/oV and soon; .2 = .20 = 
.200 = .2000, &,c. therefore the value of a decimal is not 
changed by annexing a cipher to the end of it, nor by tak- 
ing one away. 

If there be not as many figures in the numerator as there 
are ciphers in the denominator, ciphers must be put in the 
place of tenths, hundredths, <fec. thus, yf^^ is written .02 ; 
To ^s written .025 ; and,^ j3___ jg written .00003 ; and so 
on. 

Hence, the value of figures in decimals are diminished in 
the same ratio from the decimal 'point towards the rights as 
tchole numbers are increased from the right towards the left. 

When the fractional part of a mixed number is reduced 
to a decimal, the decimal part is separated from the whole 
number by a decimal point. 

Thus, 3y"?g5^ is written 3.75 ; 4yo^ is written 4.005 ; and 
so on. 

From what has been already observed, it is plain, that 
any fraction may he reduced to a decimal by adding ciphers 
to the numerator and dividing by the denominator. 

Thus, the fraction -^-^ is reduced to .05, by adding two 
ciphers to 35, and dividing the expression 35.00 by 700 ; 
as there are no tenths a cipher is put in the place of tenths ; 
so that the decimal equivalent to the fraction ^^^ is five 
hundredths. 

Again, the fraction -^-^, reduced to a decimal, is equiva- 
lent to .030303, and so on ; here, there would still be a re- 
mainder, and it is also evident that the decimal would never 



terminate ; in which case, it is only necessary in most cal- 
culations to use six or seven figures of the decimals. 

A quantity of one denomination may be reduced to the 
decimal of another quantity of the same kind., hut of a diffe- 
rent denomination ; hy first expressing the ratio of the former 
to the latter hy a common fraction, and then reducing the 
fraction thus formed to a decimal. 

For example, 2 nails is the ^^ of a yard, or i of a yard, 
which reduced to a decimal is equivalent to .126; hence, 
2 nails is the 125 thousandths of a yard. 

The reduction of a fraction to a decimal, or of one quan- 
tity to the decimal of another, is usually called reduction of 
decimals. 

Examples in Reduction of Decimals. 

Example 1. What decimal of a foot is 9 inches? 

Here, 9 inches is the ^-^ or | of a foot, which, reduced 
to a decimal, is equivalent to .75, or 75 hundredths. 

Ex. 2. What decimal of a yard is 2 feet 6 inches ? 

Here, 2 feet 6 inches is the ff or f of a yard, which, re- 
duced to a decimal, is .833333, &c. This is a repeating 
or circulating decimal, never terminating. 

Ex. 3. What decimal of an acre or 160 square poles, is 
2 roods and 16 square poles ? 

Here, 2 roods and 16 square poles is the ^^^ or f of an 
acre, which, reduced to a decimal, is .6 or 6 tenths. 

Ex. 4. What decimal of a cubic foot is 144 cubic inches ? 

Here, 144 cubic inches is the yYg^ or r^^ of a cubic foot ; 
and Jg-, reduced to a decimal, is equivalent to .083333, &c. 
being a repeating decimal. 

Ex. 5. Reduce 8 feet 6 inches to the decimal of a mile. 

Answer, .0016098. 

Ex. 6. Reduce 2 feet 5 inches to the decimal of a yard. 

Ans. .805555. 

Ex. 7. Reduce 5i yards to the decimal of a mile or 1760 
yards. Ans. .003125. 

Ex. 8. Reduce A\ miles to the decimal of 40 miles. 

Ans. .1125. 

Ex. 9. Reduce 3 roods 1 1 poles to the decimal of an 
acre. Ans. .81875. 

The decimal of one denomination may he reduced to whole 
numbers of lower denominations, as in the reduction of quan- 



titles of a higher denomination to a lower, observing, after 
each multiplication^ to point off for decimals as many figures 
towards the right as there were figures in the given decimal. 
The figures on the left hand of the decimal points will be 
the whole numbers required. 

For example, ,3945 of a day is equal to the fraction 
t VoVo ^^^ ^^Y' which, expressed as the fraction of an hour, 
is tWoV X 24=f A6^, or 9 hours and -j^^^o of an hour ; 
but Vwo%- of 3" hour is ^%%%''q X 60 of a minute, or 
VoVoV ? which is equal 28 minutes, y ^ggg of a minute; 
again, this fraction of a minute is equal to y ^ggp X 60 of 
a second, or f^wo» which is equal to 4 seconds, and ^\ of 
a second ; so that the decimal .3945 of a day is equal to 9 
hours, 28 minutes, 4j\ seconds. From this it appears that 
pointing off the decimals serves the same purpose as divid- 
ing by the denominator : thus, 
. ' .3945 day 

24 



15780 
7890 



9.4680 hours. 
60 



28.0800 minutes. 
Examples in finding the values of Decimals. 
Ex. 1> Required the value of .375 of a yard. 

Aus. 1 qr. 2 nails. 
Ex. 2. Required the value of .625 of an acre. 

Ans. 2 roods 20 poles. 
Ex. 3. Required the value of .875 of a mile. 

Ans. 7 furlongs. 
Ex. 4. Required the value of .2385 of a degree. 

Ans. 14' 18" 36 thirds. 

x\DDITION OF DECIMALS. 

The addition of decimals is performed like that of whole 
numbers, observing however to arrange the numbers so that 
the separating points may be in the same column ; that is, 
the tenths under tenths, the hundredths under hundredths, 
and so on. 



DECIMALS. 5 

For instance, the decimals .571, .672, .3, .003, and 
.0075, being arranged as follows: 
.571 
.672 
.3 

.003 
.0075 



1.5535 
their sum is found to be 1.5535; the reason of the arrange- 
raent is evident, since those figures are added together 
which are of the same local value. 

Again, the sum of the numbers 3.5, 7.005, 4.325, .0003, 
and 1.000007, which contain whole units, is found in like 
manner, thus: 

3.5 
7.005 
4.325 
.0003 
1.000007 



Sum 15.830307 
Examples in Addition of Decimals, 

Ex. 1. Required the sum of 5.714, 3.456, .543, and 
17.4957. Ans. 27.2087. 

Ex. 2. Required the sum of 3.754, 47.5, .00857, and 
37.5. Ans. 88.76257. 

Ex. 3. Required the sum of 54.34, .375, 14.795, and 
1.5. Ans. 71.01. 

Ex. 4. Required the sum of 37.5, 43.75, 56.25, and 
87.5. Ans. 225. 

Ex. 5. Required the sum of .375, .625, .0625, .1875, 
.3125, .4375, .005, .9475, and .0075. An$. 2.96. 

The Subtraction of Decimals is performed in the same 
manner as that of ivhole numbers ; observing to place each 
.figure of the less below a figure of the same local value in 
the greater. 

For instance, let the difference .of .3765 and .1236 be 
required: the decimals being arranged thus: 
A 2 



.3765 
.1236 

.2529 
their difference will be .2529. 

Again, let .7562 be taken from .82; by annexing ciphers 
to the greater and arranging the numbers thus: 

.8200 
.7562 



.0638 
we shall find the difference to be .0638: it must be ob- 
served, that the value of a decimal is not increased nor 
decreased by annexing ciphers to it; for a fraction does not 
alter its value by annexing ciphers to its numerator and 
denominator, thus; j^o = tVoV = tVoVo? ^"^ ^^ o"- "^^^^ 
is also evident from the decimal notation, which is similar 
to that of whole numbers; that is, the value of the decimal 
.82 is 8 tenths and 2 hundredths, the value of the decimal 
.820 is also the same, being 8 tenths 2 hundredths and 
thousandths; and so on. 

Examples in Subtraction of Decimals. 

Ex. 1. Required the difference between 57.49 and 5.768. 

Ans. 51.722. 



Ex. 2. 
.00075. 


Required 


the 


difference 


between .0076 and 
Ans. .00685. 


Ex. 3. 

1.2591. 


Required 


the 


difference 


between 3.468 and 
Ans. 2.2089. 


Ex. 4. 

.5236. 


Required 


the 


difference 


between 3.1416 and 
Ans. 2.6180. 



From the multiplication of fractions, (or even the deci- 
mal notation,) it appears evident, that the multiplication of 
decimals is performed as in wliole numbers, but if there be 
not as many decimals in the product as there are in both fac- 
tors^ ciphers must be prefixed to supply the deficiency. 

For instance, the product of .06 x .004 is equal to 
.00024; since .06 is equal yf g-, and .004 is equal y^^^ ; 
hence, ^^ x y^ = yof^, which, expressed accord- 
ing to the decimal notation, is equal to .00024. 



DECIMALS. 7 

Examples in Multiplication of Decimals. 

Ex. 1. Multiply 3.125 by 2.75. Ans. 8.59376. 

Ex. 2. Multiply 79.25 by .459. Ans. 36.37575. 

Ex. 3. Multiply .135272 by .00425. Ans. .000574906. 

Ex. 4. Multiply .004735 by .0375. A«5. .0001775625. 

The Division of Decimals is performed in the same man- 
ner as that of whole numbers, but the dividend must contain 
as many decimal figures as the divisor, if not ^ ciphers must 
he annexed ; and the decimals in the quotient, must be always 
equal to the excess of the decimal figures in the dividend above 
those in the divisor, if not, ciphers must he prefixed. 

For instance, when the denominators of any two frac- 
tions are the same, their quotients are found by dividing: 
their numerators: thus, 1^%-^ H- y-^ is equal to 25 -r- 5; 
that is 5: hence .025-i-.005 is equal to 5; that is, a whole 
number. 

Again, ^\ ^ j^ is equal to -^^-^ -^ yoVo; which is 
equal to 3|o_.75^ hence to the decimal .3, the dividend, 
two ciphers must be added, in orcier to have as many deci- 
mal places as the divisor .004, before the division can be 
performed. 

It likewise follows, that tV^ -i- j^-q is equal to ^^^ -r- 
T^o%V' which is equal to |^; this by reduction is equal to 
3^^, which may be written .75; hence, .375 ~ .5 = .75; 
that is, the decimal figure in the quotient is equal to the 
excess of the decimal figures in the dividend above those 
in the divisor. 

Examples in Division of Decimals. 

Ex. 1. Divide .1342 by 67.1. Ans. .002. 

Ex. 2. Divide 1.7144 by 1.5. Ans. 1.142955. 

Ex. 3. Divide 24880 by 360. A«5. 69.11 1, &c. or 69i. 

Ex. 4. Divide 172.8 by .144. Ans. 1200. 

Ex. 5. Divide .88 by 88. Ans. 100. 

Ex. 6. When the diameter of a circle is 1 the circumfe- 
rence is 3.14159 nearly; what is the diameter of the earth, 
allowing its circumference to be 24880 miles? 

Ans. 7919.53666 miles, nearly 

Extraction of the Square Root. 
The square of the sum of two numbers is equal to the 
squares of the numbers with twice their product. Thus, the 



8 EXTRACTION OF 

square of 24 is equal to the squares of 20 and 4 witli twice 
the product of 20 and 4; that is, to 400 + 2 X 20 X 4 -f 
16 = 576. Here in extracting the second root of 576, we 
separate it into two parts, 500 and 76. Thus, 500 contains 
400, the square of 20, with the remainder 100; the first 
part of the root is therefore 20, and the remainder 100+76, 
or 176. 

Now, according to the principle above mentioned, this 
remainder must be twice the product of 20, and the part of 
the roots still to be found, together with the square of that 
part. Now, dividing 176 by 40, the double of 20, we find 
for quotient 4; then this part being added to 40, the sum is 
44, which being multiphed by 4, the product 176, is evi- 
dently twice the product of 20 and 4, together with the 
square of 4. The operation may, in every case, be illus- 
trated in the same manner. Hence the following rule for 
extracting the square root of any number. 

Commencing at the unit figure, cut off periods of two 
figures each, till all the figures are exhausted, the first figure 
of the square rootunll be the square root of the first period, 
or of the greatest square contained in it, if it he not a square 
itself Subtract the square of this figure from the first pe- 
7'iod ; to the remainder annex the next period for a dividend ; 
and, for part of a divisor, double the part of the root already 
obtained. Try how often this part of the divisor is contained 
in the dividend wanting the last figure, and annex the figure 
thus found to the parts of the root and of the divisor already 
determined. Then multiply and subtract as in division; to the 
remainder bring down the next period ; and, adding to the 
divisor the figure of the root last found, proceed as before. 

For instance, the square root of 106929, is found thus: 
Square. Root. 
106929 I 327 



62)169 
124 



647)4529 
4529 



THE SQTJARE ROOT. 9 

If any thing remain, after containing the process till all 
the figures in the given number have been used, proceed 
in the same manner to find decimals, adding, to find each 
figure, two ciphers. 

If the root of a fraction be required, let the fraction be 
reduced to a decimal, and then proceed as in the extraction 
of the roots of whole numbers. 

Examples in extracting the Square Root 
Ex. 1. Required the square root of 2^ or 2.25. 

Ansicer^ 1.5. 
Ex. 2. Required tlie square root of 152399025. 

Ans. 12345. 
Ex. 3. Required the square root of 5499025. 

Ans. 2345. 
Ex. 4. Required the square root of 36372961. 

Ans, 6031. 
Ex. 5. Required the square root of 10.4976. 

Ans, 3.24. 
Ex. 6. Required the square root of 9980.01. 

Ans, 99.9. 
Ex. 7. Required the square root of 2. 

Ans. 1.414213, nearly. 

Extraction of the Third or Cube Moot, 
The cube or third power of the sum of two numbers is 
equal to the cubes of the numbers increased by 300 times 
the square of the first number multiplied by the second, and 
also increased by 30 times the first multiplied by the square 
of the second, thus: 

^^ + ^ I Multiplied, 



Multiplied \ 



20x20+4X20 

+ 4X20+16 

20x20 + 2 X4X20+16=2d power. 

20 + 4 



20X20X20 + 2X4X20X20 + 20X 16 

4X20x20+2x20x16 + 64 

Third power==8000+3 x 4x20 x 20 + 3x20x 16+64 
or 8000 + 300X4x4 + 30x2x16 + 64. 



10 EXTRACTION OF 

Hence, this rule for extracting the third or cube root of 
any given number: — Commencing at the unit figure, cut off 
periods of three figures each till all the figures of the given 
number are exhausted. Then find the greatest cube number 
contained in the first period, and place the cube root of it in 
the quotient. Subtract its cube from the first period and 
bring down the next three figures ; divide the number thus 
brought down by 300 times the square of the first figure of 
the root and it will give the second figure ; add 300 times the 
square of the first figure, 30 times the product of the first 
and second figures, and the square of the second figure to- 
gether, for a divisor ; then multiply this divisor by the se- 
cond figure, and subtract the result from the dividend, and 
then bring down the next period^ and so proceed till all the 
periods are brought down. 

For instance, in finding the cube root of 48228544, the 
operation will stand thus : 

48'228'544(364 root. 

27 



3276)21228 
19656 

393136)1572544 
1572544 



Divided by 300x3^=2700 Divided by 36^x300=288800 

30X3X6=540 30X36X4= 4320 

6x6^ 36 4X4= 16 



1 St divisor 3276 2d divisor 393 1 36 



If any thing remains, add three ciphers, and proceed as 
before; but for every three ciphers that are added, one 
decimal figure must be cut of in the root. And if the cube 
root of a fraction or a mixed number be required, reduce 
the fraction to a decimal, and proceed as in whole numbers: 
the decimal part however must consist of periods of three 
figures each, if not, ciphers must be added. 



THE CUBE ROOT. 1 1 

Examples in extracting the Cube Root* 

Ex. 1. Required the cube root of 512000000. 

Ansiver, 800. 

Ex. 2. Required the cube root of 447697126. 

Arts. 765. 
Ex. 3. Required the cube root of 2. 

Ans. 1.259921. 
Ex. 4. Required the cube root of 44361864. 

Ans. 354. 

Ex. 5. Required the cube root of .0001357. 

Ans. .05138, &c. 

Ex. 6. Required the cube root of ^6" oi" '^^81 15942. 

Ans. .262 nearly. 

Ex. 7. Required the cube root of 13|. 

Ans. 2.3908. 

DUODECIMALS. 

Fractions whose denominators are 12, 144, 1728, &c. 
are called duodecimals ; and the division and subdivision 
of the integers are understood without being expressed as 
in decimals. The method of operating by this class of 
fractions, is principally in use among artificers, in com- 
puting the contents of work, of which the dimensions were 
taken mfeet, inches^ and twelfths of an inch. 

Rule. Set down the two dimensions to be multiplied to- 
gether, one under the other, so that feet shall stand under 
feet, inches under inches, &;c. Multiply each term of the 
multiplicand beginning at the lowest, by the feet in the mul- 
tiplier, and set the result of each immediately under its 
corresponding term, observing to carry 1 for every 12, 
from the inches to the ^eet. In like manner, multiply all 
the multiplicand by the inches of the multiplier, and then 
by the twelfth parts, setting the result of each term one 
place removed to the right hand when the multiplier is 
inches, and two places when the parts become the multi- 
plier. The sum of these partial products will be the answer. 

Or, instead of multiplying by the inches, &;c. take such 
parts of the multiplicand as these are of a foot. 



12 DUODECIMALS. 

Or, reduce the inches and parts to the decimal of a foot, 
and proceed as in the multiplication of decimals. 

For example, multiply 2 feet 6 inches by 2 feet 3 inches. 

2f. 6i. or, 2f. 6i. 

2 3 2 



5 

7 6 


5 

3=i n 


5 7 6 


5 7^ 



Here, the 7, which stands in the second place, does not 
denote square inches, but rectangles of an inch broad and 
a foot long, which are to be added to the square inches in 
the third place; so that, 7 X 12 -{- 6 = 90 are the square 
inches, and the product is 5 square feet, 90 square inches. 
And this manner of estimating the inches must be observed 
in all cases where two dimensions in feet and inches are 
thus multiplied together. 

Or, the product may be found by reducing the inches to 
the decimal of a foot: thus 6 inches=.5 of a foot; hence, 
2.5x2.25=5,625 square feet, but .625 of a square foot is 
equal to .625x144=90 square inches, the same as before. 

Examples in Duodecimals. 

Ex. 1. Multiply 35 feet 4^ inches into 12 feet 3^ inches. 
Ans. 434 square feet 47 square inches. 

Ex. 2. Multiply 7 feet 9 inches by 3 feet 6 inches. 

Ans. 27 square feet 18 square inches. 

Ex. 3. Multiply 7 feet 5 inches 9 parts by 3 feet 5 inches 
3 parts. Ans. 25 square feet 102| square inches. 

Ex. 4. Multiply 75 feet 9 inches by 17 feet 7 inches. 

Ans. 1331 square feet 135 square inches. 

Ex. 5. Multiply 97 feet 8 inches by 8 feet 9 inches. 

Ans. 854 square feet 84 square inches. 



PRACTICAL GEOMETRY 



DEFINITIONS. 

1. Geometry is that science which treats of the descrip- 
tions and properties of magnitudes in general. 

2. A 'point is that which has position, but not magni- 
tude. 

3. A line is length witliout breadth ; and its bounds or 
extremes are points. 

4. A right line is that which lies evenly between its ex- 
treme points. 



5. A superficies is that which has length and breadth 
only ; and its bounds or extremes are lines. 



6. A plane superficies is that which touches in every 
part any right line that can be drawn in that superficies. 

7. A solid is that which has length, breadth, and thick- 
ness ; and its bounds or extremes are superficies. 



zE 



14 PRACTICAL GEOSIETRr. 

8. A plane rectilineal angle is the inclination or open- 
ing of two right lines which meet in a point. 



9. One line is said to be perpendicular to another, when 
it makes the angles on both sides of it equal to each other. 



10. A right angle is that v/hich is formed by two lines 
that are perpendicular to each other.* 



11, An acute angle is that which is less than a right 
angle. 



12. An obtuse angle is that which is greater than a right 
anffle. 



* Anj ang^Ie differing from a right one, is, bj some writers, called 
a oblique angles 



FBACTIC-^JL GEQ3IETEY* 



15 



13. A circle is a plane figure, formed by the revolution 
of a right line about one of its extrenaities, which remains 
fixed.* 




14. The centre of a circle is the point about which it is 
described ; and the circumference is the line or boundary 
by which it is contained. 

15. The radius of a circle is a right line drawn firorn the 
centre to the circumference. 




16. The diameter of a circle is a right line passing- 
through the centre, and terminated both ways by the cir- 
cumference. 




* N. B. The circumference itself, for the Bake of conciseness, is 
sometimes called a circle. 



I 



16 PRACTICAL GEOMETRY. 

17. An arc of a circle is any part of its periphery or cir- 
cumference. 



18. A chord is a right line joining the extremities of an 



19. A segment of a circle is any part of a circle bounded 
by an arc and its chord. 




20. A sector is any part of a circle bounded by an arc 
and its two radii drawn to its extremities. 

N. B. A semicircle is half a circle, and a quadrant the 
quarter of it. 




PRACTICAL GEOMETRY. 



17 



21. A zone is a part of a circle included between two 
parallel chords and their intercepted arcs. 




22. A sine of an arc is a right line drawn from one ex- 
tremity of an arc perpendicular to a diameter passing through 
the other extremity, as AB. 

23. The vened sine of an arc is that part of the diame- 
ter which is intercepted between the sine and the arc ; as 
BD or BE. 

24. The cosine of an arc, is that part of the diameter in- 
tercepted between the sine and centre, as CB, and is al- 
ways equal to the difference between the versed sine and 
the radius. 




Note. The heig-ht of a segment is the part of a diameter 
contained betweea the middle of the chord and the arc : 
and the difference between this and the radius is sometimes 
called the central distance. 

25. All plane figures bounded by three right lines are 
called triaJigles. 

26. An equilateral triangle is that whose three sides are 
all equal. 




18 



PRACTICAL GEOMETRY. 



27. An isosceles triangle is that which has only two 
of its sides equal. 




28. A scalene triangle is that which has all its three , 
sides unequal. 




29. A right-angled triangle is that which has one right 
angle.* 




30. An obtuse-angled triangle is that which has one ob- 
tuse angle. 




31. An acnte-angled triangle is that which has all its 
angles acute. 

* Any triang-Li differbig^ from a right-angled one is called an ohliqut- 
mngled triangle. 



PRACTICAL GEOMETRY. 



19 




32. All plane figures, bounded by four right lines, are 
called quadrangles or quadrilaterals. 

33. A square is a quadrilateral, whose sides are all equal, 
and its angles all right angles. 



34. A rhombus is a quadrilateral, whose sides are all 
equal, but its angles not right angles.* 



35. Parallel right lines are such as are in the same 
plane, and which being produced ever so far both ways, do 
not meet. 



36. A parallelogram is a quadrilateral whose opposite 
sides are parallel. 



* This figure, by working mechanics, is sometimes called a lozenge. 



20 rRACnCAL GEO^rETRY. 

37. A rectangle is a parallelogram whose angles are all 
right angles. 



38. A rhomboid is a parallelogram whose angles are not 
right angles. 



39. All other four-sided figures, besides these, are called 
trapezhtms. 

40. A right line joining any two opposite angles of a 
four-sided figure is called the diagonal. 



\ 




41. All plane figures contained under more than four 
sides are called polygons. 

42. Polygons having five sides, are called pentagons; 
those of six sides, hexagons; those of seven, heptagons; 
and so on. 

43. A regular polygon is that whose angles as well as 
sides are all equal. 

44. The base of any figure is that side on which it is 
supposed to stand; and Xhe altitude is the perpendicular 
falling upon it from the opposite angle. 

45. In a right-angled triangle the side opposite to the 
right angle is called the hypothemise ; and the other two 
sides are called legs. 

46. An angle is usually denoted by three letters, the one 
which stands at the angular point being always to be read 
in the middle. 



PRACTICAL GEOMETRY. 
C D 



21 




*47. The circumference of every circle is supposed to 
be divided into 360 equal parts, called degrees ; each de- 
gree into 60 equal parts, called minutes; and so on. 

48. The measure of any right-lined angle is an arc of a 
circle contained between the two lines which form that 
angle, the angular point being the centre. 




Note. The angle is estimated by the number of degreeg 
contained in the arc ; whence a right angle is an angle of 
90 degrees, or \ of the circumfereno^- 

PROBLEM I.f 

To divide a given line AB into two equal parts* 



E? 



* This and the following definition are used only in Practical Ge- 
ometry. 

t The demonstrations of most of these problems may be found hx 
Euclid's Elements, 



22 PRACTICAL GEOMETRY* 

1. From the points A and B, as centres, with any dis- 
tance greater than half AB, describe arcs cutting each 
other in n and nu 

2. Through these points, draw the Hue wEw, and the 
point E, where it cuts AB, will be the middle of the line re- 
quired. 

PROBLEM II. 
To divide a g-iven angle ABC into two equal parts* 




1. From the point B, with any radius, describe the arc 

2. And from AC, with the same, or any other radius, 
describe arcs cutting each other in n. 

3. Through the point n draw the line B/i, and it will bi- 
sect the angle ABC, as was required. 

PROBLEM III. 

From a given point C, in a given right line AB, to erect a 
perpendicular. 

Case I. When the point is near the middle of the line. 



PRACTICAL GEO^IETEY. 23 

1. On each side of the point C take any two equal dis- 
tances C?i, Cju. 

2. From n and ?n, with any radius greater than Cn or Cm, 
describe arcs cutting each other in 5. 

3. Through the point s, draw the line sC, and it will be 
the perpendicular required. 



Case II. When the poi?U is at, or near, the end of the line* 
0/ 



.-•'C 



-B 



1. Take any point o, and with the radius or distance oC, 
describe the arc mCn, cutting AB in m and C. 

2. Through the centre o, and the point m, draw the line 
mon, cutting the arc mCn in n, 

3. From the poi^it «, draw the line wC, and it will be the 
perpendicular required. 



Another method. 
5^. 



1. From the point C, with any radius, describe the arc 
mm, cutting the line AC in r. 

2. With the same radius, and r as a centre, cross the 
arc in n ; and from n in like manner, cross it in m. 



24 



PRACTICAL GEOMETRY. 



3. From the points n and m, with the same, or any other 
radius, describe arcs cutting each other in s. 

4. Through the point s, draw the line sC, and it will be 
the perpendicular required.* 

PROBLEM IV. 

From a given point C, oi/t of a given line AB, to let fall a 
perpendicular. 

Case I. When the point is nearly apposite to the middle 
of the line. 



Cr 



1. From the point C, with any radius, describe the arc 
nm, cutting AB in n and m. 

2. From the points n, m, with the same or any other ra- 
dius, describe two arcs cutting each other in s. 

3. Through the points C, 5, draw the line CGs, and CG 
will be the perpendicular required. 

Case II. Wlien the point is nearly opposite to the end 
of the line. 



* For another method of raising a perpendicular from any point in 
a given line, see Prob. XXXIX. 



PKACTICAL GEOMETRY. 



25 



'G 



1. To any point m, in the line AB, draw the line Cm. 

2. Bisect the line Cm, or divide it into two equal parts 
in the point n. 

3. From n, with the radius nm, or nC, describe the arc 
CGm, cutting AB in G. 

4. Through the point C, draw the line CG, and it will 
be the perpendicular required. 



Another method* 



B i 



• ri 



1. From A, or any other point in AB, with the radius 
AC, describe the arc CD. 

2. And from any other point n, in AB, with the radius 
nC, describe another arc cutting the former in C, D. 

3. Through the points C, D, drawn the line CGD, and 
CG will be the perpendicular required. 

N. B. Perpendiculars rnay be more easily raised, and 
C 



26 



PRACTICAL GEOMETKY. 



let fall, in practice, by means of a square, or other proper 
instrument. 

PROBLEM V. 

/^o trisect^ or divide a right angle ABC into three eqtm 
parts. 




1. From the point B, with any radiu3 BA, describe th«| 
arc. AC, cutting the legs BA, BC, in A, C. 

2. And from the point A, with the radius AB, or BC 
cross the arc AC in n. 

3. Also with the same radius, from the point C, cross it 
in m. 

4. Through the points m, n, draw the lines Bra, B/i, and 
they will trisect the angle as was required. 



PROBLEM VI. 

Ai a given point D to make an angle equal to a given 
angle ABC. 




FRACTICAL GEOMETRY. 27 

1. Prom the point .B, with any radius, describe the arc 
iw», eutting^ the legs BA,, BC, in the points m, n. 

2. Draw the line DE, and front the point D, with the 
same radius as before, describe the are rs. 

3. Take the distance miu on the former arc, and apply 
it to the arc r.?, from r to 5^. 

4. Through the points D, s, draw the line DF, and tlie 
angle EDF will be equal to the angle ABC, as was re- 
quired. 

PROBLEM VII. 
To draw a line parallel to a given line AB, 

Case I. When the parallel line is to pass through a 
given point C. 

C r 

G T— / H 



A » »t B 

1. To AB, from the point C, draw any right line Cwt. 

2. From the point m, with the radius mC, describe the 
arc Cm, cutting AB in n. 

3. And with the same radius, from the point C^ describe 
the arc ?/?r. 

4. Take the distance Ck, and apply it to the arc mr, 
from m to r. 

5. Through the points C, r, draw the line GCrH, and it 
will be parallel ito AB, as was required. 

Another methoih 

G — ?^- 1 E 



A B\ /'& ^ 



^' 



28 PRACTICAL GEOMETRY. 

1. From C, draw any line CDE, and make DE equal lo 
DC. 

2. From E, draw any line EFI, and make FI equal to 
EF. 

3. Through C, I, draw the line GCIH, and it will be 
parallel to AB. 



Case II. When the parallel line is to be at a given dis' 
tancefrom AB. 



1. From any two points r, s, in the line AB, with a ra- 
dius equal to the given distance, describe the arcs, n, to, 

2. Draw the line DG, to touch those arcs without cutting 
them, and it will be parallel to AB, as was required. 

N. B. The former case of this problem, as well as sev- 
eral other operations in Practical Geometry, may be more 
easily effected by means of the parallel ruler.* 



PROBLEM VIII. 

To divide a given line AB into any proposed number of 
equal parts. 



* This ruler may be had of all sizes, but is usually put into a porta- 
ble case, with a drawing-pen, scale, compasses, and other useful in- 
struments. 



PHACnCAL GEOMETHy. 

6 ^ - 



.t 



?. 



3....^r 






71 ^ 



••"^ 



1. From one enii. of the line A, draw Aw, making ahy 
angle with AB: and from B, the other end, draw B?i, 
making an equal angle ABw. 

2. In each of the lines Am, B/j, beginning at A and B, 
set off as many equal parts, of any length, as AB is to be di- 
vided into. 

3. Join the points A5, 1, 4, 2, 3, &c» and AB will be 
divided as was required. 

Nate. Bre may be drawn parallel to Am, by means of a 
parallel ruler. 

PROBLEM IX. 

To find the centre of a given circle^ or one alreadf 
described.* 




• The centre of a given circle, or of any axe of it, may also be found 
by joining three points in the circumference, and proceedin|^ aa m 
Prob. XXIV. 



Prob. XXIV. 

C2 



so 



PRACTICAL GEOMETRY. 



1. Draw any chord AB, and bisect it with the perpen- 
dicular CD. 

2. Bisect CD, in like manner, with the chord EF, and 
their intersection O, will be the centre required. 



PROBLEM X. 



To dram a tangent to a given circle^ that shall pa^s through 
a given point A. 

Case I. When the point A is in the circumference of the 
circle. 




I 



1 From the given point A, to the centre of the circle, 
draw the radius AO. 

2. Through the point A, draw CD perpendicular to OA, 
and it will be the tangent required. 

Casf ][. When the point A is witkovt the circle. 




PRACTICAL GEOMETRY. 31 

1. To the point A, from the centre O, draw the line OA, 
and bisect it in n, 

2. From the point w, with the radius wA, or nO, describe 
the semicircle ABO, cutting the given circle in B. 

3. Through the points A, B, draw the line AB, and it 
will be the tangent required. 



PROBLEM XL 
To two given lines, A, B, to find a third proportional. 




1. From the point C draw two right lines, making any 
angle FCG. 

2. In these lines take CE equal to the first term A, and 
CG, CD, each equal to the second term B. 

3. Join ED, and draw GF parallel to it ; and CF will be 
the third proportional required. 

That is CE(A) : CG(B) : : CD(B) : CF. 

PROBLEM XIL 

To three given right lines, A, B, C, to find a fourth 
proportional. 



A 

C DZ 




32 PRACnCAL GEOMETRY. 

1- From the point D draw two right h'nes, making any 
angle GDH. 

2. In these lines take DF equal to the first term A, DE 
equal to the second B, and DH equal to the third C. 

3. Join FE, and draw HG parallel to it, and DG will be 
the fourth proportional required. 

That is DF(A) : DE(B) : : DH(C) : DG. 



PROBLEM Xni. 

Between two given right lines A, B, to find a mean pro- 
portional. 




E O D 



1. Draw any right line, in which take CE equal to A, 
and ED equal to B. 

2. Bisect CD in O, and with OD or OC, as radius, de- 
scribe the semicircle CFD. 

3. From the point E draw EF perpendicular to CD, and 
it will be the mean proportional required. 

That is CE(A) : EF ; ; EF i ED(B). 



PROBLEM XIV. 

To divide a given line AB in the same proportion thai an- 
other given line C is divided.* 



* The case of this problem which most frequently occurs, is that in 
•which the given line is required to be divided into tv.o parts tJiat shall 
have a given ratio ; wliich may be done in nearly the same manner as 
above. 




33 



1. From the point A draw AD equal to the given line 
C, and making any angle with AB. 

2. To AD apply the several divisions of C, and join DB. 

3. Draw the lines 4 4, 3 3, &;c. each parallel to DB, and 
the line AB will be divided as was required. 

That is, the parts A 1, 1 2, 2 3, 3 4, 4B, on the line AB, 
will be proportional to the parts 1, 1 2, 2 3, 3 4, 4 5, on 
the line C. * 



PROBLEM XV. 

Upon a given right line AB, to make an equilateral triangle. 

"S. 




1. From the points A and B, with a radius equal to AB, 
describe arcs cutting in C. 

2. Draw the lines AC, BC, and the figure ACB will be 
the triangle required. 

Note. An isosceles triangle may be formed in the same 
manner, by taking any distance for radius. 






d 



34 PRACTICAL GEOMETRr. 

PROBLEM XVI. 

To maJce a triangle whose three sides shall he respectively 
equal to three given lines^ A, B, C* 




1. Draw a line DE equal io one of the given line^ C. 

2. On the point D, with a radius equal to B, describe an 
arc. 

3. And on the point E, with a radius equal to A, describe 
another arc, cutting the former in F. 

4. Draw the lines DF, EF, and DFE will be the triangle 
rec^uired. ^ 

PROBLEM X\1L 
Upon a given line AB to describe a square. 
T>L C 




1. From the point B, draw BC perpendicular, and equal 
to AB, 



* The three gvfeyi lines mnst be of such lengths tliat. any two of 
them taken together shaQ be greater than the third. 



P2ACTICAL GE03IETRY. ^ 

2. On A and C, with the radius AB, describe two arcs 
cutting each other in D. 

3. Draw the lines AD, CD, and the figure ABCD will 
be the squar^e required. 

Note. ^ A rhombus may be made on the given line AB 
in exactly the same manner, if BC be drawn with the proper 
obliquity, instead of perpendicularly. 

PROBLEM XVin. 

To describe a rectangle, whose length and hreadth shall he 
equal to two given lines AB and C. 




1. At the point B, in the given line AB, erect the per- 
pendicular BD, and make it equal to C. 

2. From the points D, A, with the radii AB and C de- 
scribe two arcs cutting each other in E. 

3. Join EA and ED, and ABDE will be the rectangle 
required. 

Note. A parallelogram may be described in nearly the 



same manner. 



PHOBLEMXIX. 
In a given triangle ABC to inscribe a circle* 
C 




36 PRACTICAL GEOMETRY. 

1. Bisect the angles A and B with the lines AO and 
BO. 

2. From the point of intersection O let fall the perpen- 
dicular On, and it will be the radius of the cixcle required. 



PROBLEM XX. 

In a given circle to inscribe an equilateral triangle, a hexa- 
gon, or a dodecagon. 



For the hexagon. 

1. From any point A as a centre, with a distance equal 
to the radius AO, describe the arc BOF. 

2. Join the points AB, or AF, and either of these lines 
being carried six times round the circle will form the hexa- 
gon required. 

That is, the radius of the circle is equal to the side of 
the hexagon ; and the sides of the hexagon divide the cir- 
cumference of the circle into six equal parts, each contain- 
ing 60 degrees. 

For the equilateral triangle. 

1. From the point A, to the second and fourth divisions, 
or angles of the hexagon, draw the lines AC, AE. 

2. Join the points CE, and ACE will be the equilateral 
triangle required ; the arc AC being one third of the cir- 
cumference, or 120 degrees. 



i 



PKACTICAL GEOMETItY- 37 

For the dodecagon. 

Bisect the arc AB of the hexaoon in the point n, and the 
line An being carried twelve times round the circumfer- 
ence, will form the dodecagon required, the arc An being 
30 degrees. 

If An be again bisected, a polygon may be formed of 24 
sides ; and by another bisection a polygon of 48 sides ; and 



PROBLEM XXI. 

To inscribe a square, or an octagon, in a given circle. 

A 




For the square. 

1. Draw the diameters BD and AC, intersecting each 
other at right angles. 

2. Join the points AB, BC, CD, and DA, and ABCD 
will be the square required. 

For the octagon. 

Bisect the arc AB of the square in the point E, and the 
line AE being carried eight times round the circumference 
will form the octagon. 

If the arc AE be again bisected, a polygon may be form- 
ed of 16 sides : and by another bisection, a polygon of S2 
sides J and so on. 
D 



39 FKACTICAL GEOMETRY. 

PROBLEM XXIL* 
To inscribe a pentagon, or decagon, in a given circle* 



P 

For the pentagon, 

1. Draw the diameters Ap, nm, at right angles to each 
other, and bisect the radius On in r. 

2. From the point r, with the distance r A, describe the 
arc As, and from the point A, with the distance As, de- 
scribe the arc sB. 

3. Join the points A, B, and the line AB being carried 
five times round the circle, will form the pentagon required. 

For the decagon. 

Bisect the arc AE of the pentagon in c, and the line Ac 
being carried ten times round the circumference will form 
the decagon required. 

If the arc AC be again bisected, a polygon of 20 sides 
may be formed ; and by another bisection a polygon of 40 
sides ; m\d so on. 



* Besides the figures here constructed, and those arising* from thence 
by continual bisections, or taking the diiFerences, no other regular pol- 
ygon caa be described, from any known mG\h.oA purely gtometricaU 



PRACTICAI* GE03tETRY. 



89 



PROBLEM XXni, 

On a given line AF, to describe a regular polygon of any 
proposed number of sides* 




1. From the point A, with the distance AF, describe 
the semicircle FBG, which divide into as many equal parts 
Ga, aB, Be, &c. as the pob/gon is to have sides.* 

2. From A to the second point of division draw AB, and 
through the other points c, d, e, dsc. draw the lines AC, 
AD, AE, &c. 

3» Apply the distance AF from F to E, from E to D, 
from D to C, &c. and join BC, CD, DE, &c. and ABCD, 
&c. will be the regular polygon required. 



PROBLEM XXIV, 

About a given triangle ABC to circumscribe a circle. 



* The semicircle is conveniently divided by means of a scale of 
chords, with the use of which the student is supposed to be ae* 
quainted. 



40 



FRACnCAL GEOMETHT, 
C 




1. Bisect the two sides AB, BC, with the perpendiculars 
moy and no, 

2. From the point of intersection o, with the distance 
OA, OB, or OC, describe the circle ACB, and it will be 
that required. 

If any two of the angles be bisected, instead of the sides, 
the intersection of the lines will also give the centre of the 
inscribed circle. 

PROBLEM XXV. 

About a given square ABCD to circumscribe a circle* 



><0 



1. Draw the two diagonals AC and BD intersecting each 
other in O. 

2. From the point O, with the distance OA, OB, OC or 
OD, describe the circle ABCD, and it will be that re- 
quired. 

PROBLEM XXVL 

To circumscribe a square about a given circle* 



FSACTICAL GEOMETRY. 



41 




I. Draw any two diameters no and rm at right angles to 
each other. 

Through the points m, o, r, «, draw the lines AB, BC, 
CD, and DA, perpendicular to rm, and wo, and ABCD will 
foe the square required.* 

PROBLEM XXVIl. 

About a given circle to circumscribe a pentagon^ 

D 




1 . Inscribe a -pentagon in the circle ; or, which is the 
same thing, iind the points m, n^ v, r, s^ as in Prob. XXII. 

2. From the centre o, to each of these points, draw the 
radii an, om, ov, or, and as. 

3. Through the points n, m, draw the lines AB, BC, 
perpendicular to on, om; producing them till they meet 
each other at B.' 

4. In the same manner, draw the lines CD, DE, EA, and 
ABCDE will be the pentagon required. 

* If each of the quadraiits rn, mn, mo and or, be bisected, and 
tangents be draviTi to those points, the circumscribing figure will be 
an octagon. 

D2 



42 PRACTICAL GEOMETRY. 

Note, — Any other polygon may be made to circumscribe 
a circle, by first inscribing a similar one, and then drawing 
tangents to the circle at the angular points. 



PROBLEM XXVm.* 
On a given line AB to maJce a regular pentagon. 








A-- ...-'B 



1. Make Bm perpendicular to AB, and equal to one 
half of it. 

2. Draw Am, and produce it till the part Tim is equal to 
Bm. 

3. From A and B as centres, with the radius Bn, de- 
scribe arcs cutting each other in o. 

4. And from the point o, with the same radius or with 
oA, or oB, describe the circle ABCDE. 

5. Apply the line AB five times round the circumference 
of this circle, and it will form the pentagon required. 

Note. — If tangents be drawn through the angular points 
A, B, C, D, E, a pentagon circumscribing the circle will 
be formed ; and if the arcs be bisected, a circumscribing 
decagon may be formed.. 



* In the former edition of this work, another method of describing 
a pentag^on was given, as first proposed by xMbertus Durer, in his Ge- 
ometry, p. 55, printed in 1532 ; but as that is only an approximation, 
and is not more easy in pi-actice than the present one, which is per- 
fectly accurate, it is here omitted. 



PRACTICAL GEOMETRY. 



PROBLEM XXIX. 



43 



On a given line AB to make a regular hexagon. 
E -.^D 



...0. 



1. From the points A, B, as centres, with the radius AB, 
describe arcs cutting each other in O. 

2. And from the point O, with the distance OA or OB, 
describe the circle ABCDEF. 

3. Apply the line AB six times round the circumference, 
and it will form the hexagon required.* 

PROBLEM XXX. 
On a given line AB to form a regular octagon. 



m A 



Ji n 



1. On the extremes of the given line AB erect the in- 
definite perpendiculars AF and BE. ^ 

2. Produce AB both ways to m and n, and bisect the 
angles mAF and tjBE with the lines AH and BC. 



* This construction is founded on the principle, that the radius of 
every circle is equal to the side of the inscribed hexagon, or the chord 
of 60°. 



44 PRACTICAL GEOMETRY. 

3. Make AH and BC equal to AB, and draw HG, CD, 
parallel to AF or BE, and also each equal to AB. 

4. From G, D, as centres, with a radius equal to AB, 
describe arcs crossing AF, BE, in F, and E ; and if GF, 
FE, and ED be drawn, ABCDEFGH will be the octagon 
required. 

PROBLEM XXXI. 

To make a figure similar to a given figure ABCDE. 




1. Take A5 equal to the side of the figure required, and 
from the angle A dra)v the diagonals Ac, Ad. 

2. From the points 6, c, <7, draw he, cd, de, parallel to 
BC, CD, DE, and Abcde will be similar to ABCDE. 

The same thing may also be done by making the angles 
b, c, d, e, respectively, equal to the angles B, C, D, E. 

PROBLEM XXXn. 
To make a triangle equal to a given trapezium ABCD. 




1. Draw the diagonal DB, and make CE parallel to it, 
meeting the side AB produced in E. 

2. Join the points D, E, and ADE will be the .triangle 
required. 



PBAeriCAL GEOMETRY. 



45 



PROBLEM XXXm. 

To make a triangle equal to any right lined figure^ 
ABCDEA. 




1. Produce the side AB both ways at pleasure. 

2. Draw the diagonals DA, DB, and parallel to them the 
lines EF and CG. 

3. Join the points DF, DG, and DFG will he the triangle 
required. 

And in nearly the same manner may any right lined 
figure whatever be reduced to a triangle. 

PROBLEM XXXIV.* 
To make an angle of any proposed number of degrees* 

c c 





* The line of chords made use of m the foUowing problems, is com- 
monly put upon the plane scale, and is adapted to 90 degrees or the 
fbiirth part of a circle. 

For a description of this and other instruments made use of in 
Practical Geometry, see Mr. Robertson's Treatise on such mathematu 
eal instruments as are usuallt/ put into a portable case^ 



46 PRACTICAL GEOMETRY. 

1. Take the first 60 degrees from the scale of chords, 
and from the point A, with this radius, describe the arc nm. 

2. Take the chord of the proposed number of degrees 
from the same scale, and apply it from n to m. 

3. From the point A draw the lines An and Awi, and 
they will form the angle required. 

Note. — Angles greater than 90° are usually taken at 
twice. 

PROBLEM XXXV.* 

An angle BAG hdng given, to find the number of degree* 
it contains* 




B A n B 



1. From the angular point A, with the chord of 60 de- 
grees, describe the arc «m, cutting the legs in the points 
n and m. 

2. Take the distance nm, and apply it to the scale of 
chords, and it will show the degrees required. 

Note. — When the distance nm is greater than 90°, it 
must be taken at twice, as before. 

PROBLEM XXXVI. 

In a given circle to inscribe a polygon of any proposed 
number of sides. 



* Both this and the last problem may be performed by means of a 
frotracior. 



PRACTICAL GEOMETKY. 47 




1. Divide 360° by the number of sides, 'and make an 
angle AOB, at the centre, whose measure shall be equal to 
the degrees in the quotient. 

2. Join the points AB, and apply the chord AB to the 
circimiference the given number of times, and it will form 
the polygon required. 

PROBLEM XXXVn. 

On a given line AB to form a regular polygon of any prO' 
posed number of sides* 



1. Divide 360° by the number of sides, and subtract the 
quotient from 180 degrees. 

2. Make the angles ABO and BAO each equal to half 

the difference last found. 

S. From the point of intersection O, with the distance 
OA or OB, describe a circle. 

4. Apply the chord AB to the circumference the pro- 
posed number of times, and it will form the polygon re- 
quired.* 

* By this method the circumference of a circle may also be divi- 
ded into any number of equal parts ; for if 360 be divided by the 



48 PRACTICAL GEOMETRY. 

PROBLEM XXXVm.* 

JJ;pon a right line AB to describe a regular 'pentagon. 




1. Produce AB towards n, and at the point B make the 
perpendicular ^m equal to AB. 

2. Bisect AB in r, and from r as a centre, with the radius 
rwi, describe the arc mn, cutting AB in n. 

3. From the points A and B, with the radius An, describe 
arcs cutting each other in D. 

4. And from the points A, D, and B, D, with the radius 
AB, describe arcs cutting each other in C and E. 

5. Join BC, DC, DE, and EA, and ABCDE will be the 
pentagon required. 

This method differs but little from that of Problem 
XXVUI, and is equally easy and convenient in practice. 

PROBLEM XXXIX. 

To raise a perpendicular from any point \in a given line 
AB. 



number of parts, and the angle AOB be made equal to the degrees in 
the quotient, the arc AB will be one of the equal parts required. 

* This and the following problem were not given in the first edition 
of this work, but are now added on account of their elegance and utility. 
The second is derived from the 47th Prop. B. I. Euclid's Elements, 
and the first is proposed &r a demonstration in the Ladies* Diary for 
the year 178a 



PEACnCAL GEOMETEY. 

c 



5/ 



49 



3 JB 



1. From any scale of equal parts take a distance equal 
to 3 divisions, and set it from B to wi. 

2. And from the points B and m, with the distances 4 
and 5, taken from the same scale, describe arcs cutting 
each other in n. 

S. Through the points n, B, draw the line BC, and it 
will be the perpendicular required. 



JSxplanaiion of the characters made use of in the following 
part of this work. 

4- Is the si^ of addition. 
of subtraction. 






■ of multiplication. 

- of division. 

- of the square root. 

• of the cube root. 

• of equality. 

• of proportion. 



REMARKS. 

1. An angle in a semicircle is a right angle. 

2. All angles in the same segment of a circle are equal. 

3. Triangles that have all the three angles of the one re- 
spectively equal to the three angles of the other, are called 
equiangular triangles, or similar triangles. 



50 PRACTICAL GEOMETRY. 

4. In similar triangles the like sides, or sides opposite to 
the equal angles, are proportional. 

5. The areas or spaces of similar triangles are to each 
other as the squares of their like sides. 

6. The areas of circles are to each other as the squares 
of their diameters, radii, or circumferences. 

7. Similar figures are such as have the same number of 
sides, and the angles contained by those sides respectively 

qual. 

8. The areas of similar figures are to each other as the 
squares of their like sides. 



MENSURATION OF SUPERFICIES. 



The area of any figure is the measure of its surface, or 
the space contained within the bounds of that surface, with- 
out any regard to thickness. 

A square whose side is one inch, one foot, or one yard, • 
&e. is called the meas-uring unit, and the area or content 
of any figure is computed by the number of those squares 
contained in that figure. 

N. B. In all questions involving decimals, they are car- 
ried out to the fourth place inclusive, and then taken to the 
nearest figure ; that is, if it be found that, by extending the 
operation, the next figure would be 5, or upwards, the fourth 
decimal figure is increased by 1. The student by observ-- 
ing this rule will generally find his results to agree with 
those given in the book. 

PROBLEM I. 

To find the area of a parallelogram ; whether it he a square^ 
a rectangle, a rho?nbus, or a rhoriiboides, 

RULE.* 

Multiply the length by the perpendicular height, and the 
product will be the area. 



* Take ftiiy rectangle ABCD, and di- 
vide each of its sides respectively, into as 
many equal parts as are expres,?ed by tlie 
number of times they contain the linear 
measuring unit, arid l&t all the opposite 
points of division be connected by right 
lines. Then it is evident, that these lines 
divide the rectangle into a number of 



c 



52 



MENSURATION 



Note. — The peri:>endicular height of the parallelogram is 
equal to the area divided by the base. 



1. Required the area of the square ABCD whose side is 
5 feet 9 inches. 




Here 5 ft. 9m. =5.75; and 5.75!^= 5.75 X 5.75=33.0625 
feet=^^'^fe. Oin. 9pa.=ar€a required. 

2. Required the area of the rectangle ABCD, whose 
length AB is 13.75 chains, and breadth BC 9.5 chains. 




Bquares each equal to tlie superficial measuring unit, and that the num- 
ber of these squares, or the area of the figure, is equal to the number 
of linear measuring units in the length, as often repeated as there are 
linear measxiring units in the breadth or height, that is, equal to the 
length multiplied by the height, which is the rule. 

And since a rectangle is equal to an obhque parallelogram standing 
upon the same base, and between the same parallels, (Euc. I. 35,) the 
rule is true for Etny parallelogram in general. Q. E. D. 

Rule II. If any two sides of a parallelogram be multiplied together, 
and the product again by the natural sine of the included angle, the 
last product will give the area of the parallelogram. That is abXbc 
= nat. suie of the angle B=area. 

Note. — ^Because the angles of a square and rectangle are each 90*^, 
whose natural sine is unity, or 1, the rule in this case is the same as 
that given in the text. 



t)P STTPEKFieiES. 



Here 13.75x9.5=130.625; and 



130.625 
10 



13.0625 



ac.=13 ac. ro. 10 j>o.= area required, 

3. Required the area of the rhombus ABCD, whose 
length AB is 12 ieei 6 inches, and its height DE 9 feet 
3 inches. 




Here 12/e. 6 m,=12.5, and 9/e. 3 i7z.=r9.25. 

Whence 12.5X 9.25=115.625/e.=115 /e. 7 iw. Q pa. 
= area required,. 

4. What is the area of the rhoniboides ABCD whose 
length AB is 10.52 chains, and height DE 7.63 chains? 




Here 10.52x7.63=80.2876; and ?^:??I^ 



8.02676 



acre5=8 ac. ro, 4 po.=n7'ea required. 

5. What is the area of a square whose side is 35.25 
chains? ac. ro. po. 

Ans. 124 1 1 

6. What is the area of a square whose side is 8 feet 4 
inches? fe. in. pa. 

Ans. 69 5 4 
E2 



54 MENSURATIOPf 

7. What is the area of a rectangle whose length is 14 
feet 6 inches, and breadth 4 feet 9 inches ? fe. in. pa. 

Ans. 68 10 6 
B. Required the area of a rhombus, the length of whose 
side is 12.24 feet, and height 9.16 feet. fe. in. pa, 

Ans. 112 1 5 

9. Required the area of a rhomboides whose length is 
10.51 chains, and breadth 4.28 chains. ac. ro, pa. 

Ans. 4 1 39.7 

10. What is the area of a rhomboides whose length is 7 
feet 9 inches, and height 3 feet 6 inches ? fe. in. pa. 

Ans. 27 1 6 

1 1 . To find the area of a rectangular board, whose length 
is 12^ feet, and breadth 9 inches. Ans. 9f feet. 

PROBLEM II. 

To find the area of a triangle. 

RULE.* 

Multiply the base by the perpendicular height, and half 
the product will be the area. 

Note. — ^The perpendicular height of the triangle is equal 
to twice the area divided by the base. 

EXAMPLES. 

1. Required the area of the triangle ABC, whose base 
AB is 10 feet 9 inches, and height DC 7 feet 3 inches. 

C 




• A triarigle is half a parallelogram of the same base and altitudo 
(E«ie> I. 41,) and therefore the truth of this rule is evident. 



OF STJPERFICrES. 55 

Here 10 fe. 9 m.= 10.75, and 7/e. 3 m.=7.25 ; 

Whencel0.75x7,25='77,9S75,and11:^^— = 38.96875 

2 
feef=^S8fe. 11 in. 7^ pa.=area required. 

2. What is the area of a ti-ian^le whose base is 18 feet 
4 inches, and height 11 feet 10 inches? fe. in. pa. 

Ans. 108 5 8 

3. What is the area of a triangle whose base is 16.75 
feet, and height 6.24 feet ? fe. in. pa. 

Ans. 52 3 1 

4. Required the area of a triangle whose base is 12.25 
chains, and height 8.5 chains. ac. ro. po. 

Ans. 5 33 

5. What is the area of a triangle whose base is 20 feet, 
and height 10.25? Ans. 102.5/e. 

PROBLEM ni. 

To find the area cf a triangle whose three sides only are 

given. 

RULE.* 

1. From half the sum of the three sides subtract each 
side severally. 

* Demon. Let AC=a, ab=&, bc=c, and ad=^x ; (See 
preceding fig.) Then, since bi>=:& — ar, we shall have 
c' — {h — xfz=cjr=^o? — 3?, or c^ — b^-{-2hx — x^=za^—a^y 

from which x is found =: — IL_ by trans, and reduc- 
tion. 2^ 

But CD^ = AC^ AD^ = AC + AD X AC AD = (d 

a^ 4- &- — cV . _ a' + h^ — c' ._ 2ah + a' + lr~(^ 

2b ^ ^ 2b ^ 26 ~ 

^ 2ab — a^ — ¥ + c^ _ {a + bf — c" ^ c' — {a — )\ 

^ 2b ^ 2b 2b ' 



Whence cd = _ V a + ^6 — c'^)x{c^ — a — 6^), and the 



area i abX CD = iV{a+l^ — (^)X (c'— a — b^) 



56 MENSURATION 

2. Multiply the half sum and the three remainders con- 
tinually together, and the square root of the product will 
be the area required. 



1. Required the area of the triangle ABC, whose three 
sides BC, CA, and AB are 24, 36, and 48 chains respec- 
tively. 

C 




Here — — — it — = = 54= ^ sum of the sides : 

2 2^ 



= i-v/(a4-& + cX a+6 — cX c -\- a — h X c — a'-\- 6) 

^2 2 2 2 ^ 

which, by making s = ^x (a + fe + c) becomes = 
■s/(sXs — cXs — bxs — a) = algebraical expression for the 
rule, as was to be demonstrated. 

Cor. 1. If s be put equal toa-\-b, and d=b ui a, the rule 
is i\/(s2 — c-)x{c- — d-). 

Cor, 2. If all the sides be equal, the rule will become 
:jf a" v/ 3, or 5 a2 X 1.732 for the equilateral triangle whose 
side is a. 

Cor. 3. If the triangle be right angled, a being the 

hypothenuse, the rule will be X — — ,. or 



\ pX-\ p — a, putting p for the perimeter. 

Rule II. Any two sides of a trianfrlc being multiplied together, and 
the product again by half the natural sine of tlieir included angle, will 
give the area of the triangle. 

That is, AC X cte X nat. sine of the angle c= twice area. 



OP SrPERFICIES. 67 

Also 54 — 24=30 first diff, 54 — 36=18 second diff 
and 54 — 48=6 third diff. 

Whence V 54x30 X 18 X 6 = V 174960 = 418.282 = 
area required. 

2. Required the area of a triangle whose three sides are 
13, 14, and 15 feet. Ans. 84 square feet. 

3. How many acres are there in a triangle whose three 
sides are 49.00, 50.25 and 25.69 chains? Ans. 61.498 ac. 

4. Required the area of a right angled triangle, whose 
hypothenuse is 50, and the other two sides 30 and 40. 

Ans. 600. 

5. Required the area of an equilateral triangle, whose 
side is 25. Ans. 270.6329. 

6. Required the area of an isosceles triangle, whose base 
is 20, and each of its equal sides 15. Ans. 111.803. 

7. Required the area of a triangle, whose three sides are 
20, 30, and 40 chains. Ans. 29 ac.7.58 jpo. 

PROBLEM IV. 

Any two sides of a right angled triangle being given to _ 
find the third side. 

RULE.* 

1. When the two legs are given to find the hypotJienuse. 
Add the square of one of the legs to the square of the 

other, and the square root of the sum will be equal to the 
hypothenuse. 

2. When the hypothenuse and one of the legs are given 

to find the other leg. 

From the square of the hypothenuse take the square of 
the given leg, and the square root of the remainder will be 
equal to the other leg. 

* By Euc. 47.Lab2 -f bc^ =ac2 ,or ac^ — ab^ =bc2 ; and 
therefore y/ xb^ -\- bc^ = ac, or v/ac^ — ab^ = bc, or 
^ AC^ — BC^ =:ab, which is the same as the rule* 



68 MENSURATION 



EXAMPLES. 



1. In the right angled triangle ABC, the base AB is 56, 
and the perpendicular* BC 33 ; what is the hypothenuse 1 




Here 562 + 332 = 3136 + 1089=4225; a?i£fv'4225=65 
=::hypofhemise AC. 

2. If the hypothenuse AC be 53, and the base AB 45 • 
what is the perpendicular BC ? 

Here 532—452=2809—2025=784; and V784=28-- - 
perpendicular BC. 

3. The base of a right angled triangle is 77, and the 
perpendicular 36 : what is the hypothenuse ? Ans. 85. 

4. The hypothenuse of a right angled triangle is 109, 
and the perpendicular 60 : what is the base? Ans. 91. 

5. It is required to find the length of a shore, which 
strutting 12 feet from the upright of a building, will support 
a jamb 20 feet from the ground. Ans. 23.3238yee^ 

6. The height of a precipice, standing close by the side 
of a river, is 103 feet, and a line of 320 feet will reach 
from the top of it to the opposite bank : required the breadi^h 
of the river. Ans. 302.9703/ee/. 

7. A ladder 50 feet long, being placed in a street, reached 
a window 28 feet from the ground, on one side ; and by turn- 
ing it over, without removing the foot, it reached another 
window, 36 feet high, on the other side : required the 
breadth of the street. Ans. 76.1233/ec< 



OP SUPERFICIES. 



59 



PROBLEM V. 

To find the area of a trapezium, 

RULE.* 

Multiply the diagonal by the sum of the two perpendicu- 
lars falling upon it from the opposite angles, and half the 
product will be the area. 

EXAMPLES. 

1. Required the area of the trapezium BAED, whose 
diagonal BE is 84, the perpendicular AC 21, and DF 28. 




4116 



Here 28 + 21X84=49x84=4116; and 3i±l=:2058 
the area reqidred. 



BE X DP 

* Demon, The area of the triangle bde is= ^ — » ^^^ 

the area of the triangle bae is= ; and therefore the 

sum of these areas, or the area of the whole trapezium, 
beXdf beXac dp + ac 
+ 5 = - 



IS: 



2 



- Xbe. 



Q. E. D. 



If the trapezium can be inscribed in a circle ; that is, if the sum of 
two of its opposite angles is equal to two right angles, or 180°, the 
area may be found thus : 

Rule. From half the sum of the four sides subtract each side seve- 
rally; then multiply the four remainders continually together, and 
the square root of the product will be the area. 



60 



ITENSUHATION 



2. Required the area of a trapezium whose diagonal is 
80.5, and the two perpendiculars 24.5 and 30.1. 

Ans. 2197.65. 

3. What is the area of a trapezium whose diagonal is 
108 feet 6 inches, and the perpendiculars 56 feet 3 inches, 
and 60 feet 9 inches? Ans. 6347 /€. 3 in 

PROBLEM VI. 

To Jind the area of a trapezoid, or a quadrangle^ two of 
whose opposite sides are parallel. 

RULE.* 

Multiply the sum of the parallel sides by the perpendicu- 
lar distance between them, and half the product will be the 
area. 

EXAMPLES. 

1. Required the area of the trapezoid ABCD, whose 
sides AB and DC are 321.51 and 214.24, and perpendicular 
DE 171.16. 

D C n 




AB X DE 

* Demoju The A abd is = — « — ' ^"" *"® ^ bcd= 

DG X B« .V . HCXDB rjyt r 

, or, (because bw=de) = . Therefore, A 



ABD X ABCD, or the whole trapezoid abcd, is = 

dcXde ab + dc 
X — ^ — = — ^ — X DE. Q. E. D. 



OP STPEHPICIES. 61 

./^''',^J^^'^^ + ^^^''^^=^^^'75=sum of the parallel 
sides AB, DC. 

me/ice 535.75 X 171.16 (tkeperp. be) =91698.9700. And 

91698.9700 ^-o 

=4o849.4S5 the area required, 

2. The parallel sides of a trapezoid are 12.41 and 8.22 
chains, and the perpendicular distance 5.15 chains; re- 
quired the area. ^^. ^^^ ^^ 

Ans. 5 1 9.956. 

3. Required the area of a trapezoid whose parallel sides 
are 25 feet 6 inches and 18 feet 9' inches, and the perpen- 
dicular distance 10 i'eet 5 inches. fe. in, pa. 

Ans. 230 5 7 

"^'oil^^^'f^ ^^'® ^^^""^ ^ trapezoid whose parallel sides 
ara 20.5 and 12.25, and perpendicular distance 10.75. 

Ans. 176.03125. 

PROBLEM Vn. 

To fold the area of a regular polygon, « 
RULE.* 

Multiply half the perimeter of the figure hj the perpen- 
dicular falling from its centre upon one of the sides, and 
the product will be the area. 
«id^''^^* ^^^ perimeter of any iigure is the sum of ail its 



* Demxtn, E^ery, regular polygon Is eomposed of as many equal 
triangles as it has sides, consequently tlie area of one of thLrS 
angles bein| multiplied by the nrnnbe. of sides must gi^e the rr;a of 
the wnole figure ; but the area of either of tiie triangles is eoual to 
the rect^gle of the perpendicular and half the base, and tberetbre t£ 
rectangle ot the perpendicular ^d half the sum of the sides is equal 
to the area of the whole polygon ; thus, ^ 

^^ ' 6ab 

opX--=area of the a aob, and opx=-=area of th^ 

^ 2 

polygon AJBCDE. Q. E. D 

F 



62 JTENSURATIO:?? 

1. Required the area of the regular pentagon ABCDE, 
whose side AB, or BC, &c. is 25 feet, and perpendicular 
OP 17.2 feet. 

D 



Here?^^=G2.d=half perimeter; and 62.5 X 17.2= 
2 
1075 square feet= area required. 

2. Required the area of a hexagon whose side is 14.6 
feet, and perpendicular 12.64. 

Ans. 553.632 square feet. 

3. Required the area of a heptagon whose side is 19.38, 
and perpendicular from the centre 20. 

Ans. 1356.6. 

4. Required the area of an octagon whose side is 9.941, 
and perpendicular 12. Ans. 477. 16S. 

PROBLEM MIL 

To find the area of a regular jyolygon, when the side only 
is given. 

RULE.* 

Multiply the square of the side of the polygon by the 
number standing opposite to its name in the following 
table, and the product will be the area. 



* Demon. The multipliers in the table are the areas of the poly- 
gons to which they belong- when tlie side is unity, or 1. 

Whence as all regular polygons, of the same number of sides, are 



OF SUPERFICIES, 



63 



No. of 
sides. 


Names. 


Multipliers. 


3 


Trigon or equil. A 


0.433013— 


4 


Tetragon or square 


1.000000 + 


5 


Pentagon 


1,720477 + 


6 


Hexagon 


2.598076 + 


7 


Heptagon 


3.633912 + 


8 


Octagon 


4.828427 + 


9 


Nonagon 


6.181824 + 


10 


Decagon 


7.694209— 


11 


Undecagon 


9.365640— 


12 


Duodecagon 


11.196152 + 



similar to each other, and as similar figures are as the squares of theii 
like sides, (^Euc. VI. 20,) 1^ : multiplier in the table : : square of the 
side of any polygon : area of that polygon ; or which is the same 
thing, the square of the side of any polygon X by its tabular number 
is =area of the polygon. Q. E. I). 

The table is formed by trigonoraetry, thus : As radius = 1 ^tang. 
BP X tan. Z. OBP 

^ OBP : : BP (A) : Po=:— — . = J tang, c obp : 

radius 
whence op X BP = Jtang. z.obp = area of the A aob ; and | tajig. 
L OBP X number of sides = tabular nmiiber, or the area of the poly, 
gon. 

The angle obp, together with its tangent, for any polygon, of not 
more than 12 sides, is shown in the following table. 



No. of 
sides. 


Names. 


Aiigle 

OEP. 


Tangents, 


3 
4 

5 

6 

7 
8 
9 

10 
11 
12 


Trigon 
Tetragon 

Pentagon 
Hexagon 
Heptagon 
Octagon ^ 
Nonagon"^ 

Decagon 

Undecagon 

Duodecagon 


30^ 

45° 

54° 
60° 

0-1'"=- 
67^4 
7C° 

72° 
75° 


..57735+ =1^3 
1.00000 + = 1x1 


1.37638 + =:Vl + iv^5 
1.73205+ = >/3 

2.07652 + 
2.41421+=l+v^2 

2.74747 + 


3.07768+ = ^/5 + 2 v^5 
3.40568 + 
3.73205+— 2+V3 



64 MENSUHATIOW 

EXAJIFLES. 

1. Required the area of a pentagon whose side is 15. 

The numher opposite pentagon in the table is 1.720477. 

Hence 1.720477 X 15^ = 1.720477 X 225=387.107325== 
ai'ea required. 

2. The side of a hexagon is 5 feet 4 inches ; what is 
the area? Ans. 73.9. 

3. Required the area of an octagon whose side is 16. 

Ans. 1236.0773. 

4. Required the area of a decagon whose side is 20.5. 

Ans. 3233.4913. 

5. Required the area of a nonagon whose side is 36. 

Ans. 8011.6439. 

6. Required the area of a duodecagon whose side is 125. 

Ans. 174939.875. 

PROBLEM IX. 

TTie diameter of a circle being given^ to find the circum- 
Jerence ; or^ the circumference being given, to find the di- 
ameter. 

RULE.* 

Multiply the diameter by 3.1416, and the product will 
be the circumference, or 



* 1'he proportion of the diameter of a circle to its circumference 
has nev^r yet been exactly ascertained. Nor can a square or any 
other right lined fig-xire, be found, that shall be equal to a given circle. 
This is the celebrated problem called the squaring of the circle, which 
has exercised the abilities of tlie greatest mathematicians for ages. 



OF SUPERFICIES* 6^5 

Divide the circumference by 3.1418, and the quotient 
will be the diameter. 

Note 1. — As 7 is to 22, so is the diameter to the circum- 
ference ; or as 22 is to 7, so is the circumference to the di- 
ameter. 

2. As 113 is to 355, so is the diameter to the circum- 
ference ; or, as 352 is to 115, so is the circumference to 

the diameter. 



and been the occasion of so many disputes. Several persons of con- 
siderable eminence, have, at different times, pretended that they had 
discovered the exact quadrature ; but tlieir error" have so»n been de- 
tected, and it is now generally looked ujion as a thing impossible to be 
done. 

But though the relation between the diameter and circumference 
cannot be accurately expressed in known numbers, it may yet be ap- 
proximated to any assigned degree of exactness. And in this manner 
was the problem solved by the great Archimedes, about two thousand 
years ago, who discovered tlie proportion to be nearly as 7 to 22, 
which is the same as our first note. Tliis he effected by showing that 
the perimeter of a circumscribed regular potygon of 192 sides, is to 
tlie diameter in a less ratio than that of 3^^ to 1, and that the perime- 
ter of an inscribed polygon of 86 sides is to the diameter in a greater 
ratio than tliat of Z\j to 1, and from thence inferred the ratio above 
mentioned, as may be seen in his book De Dimensione Circuli. The 
-same proportion was also discovered by PhiloGedarensis and ApoUo- 
nius Pergeus at a still earlier period, as we are informed by Eutocius 
in his observations on a Avork called Ocyteboos. 

The proportion of Vieta and ]\Ietius is that of ] 13 to 355, which 
is soraethLag more exact than tlie former, and is the same as the second 
note. 

This is a very commodious proportion : for being reduced into deci- 
mals, it agrees with the truth as far as the sixth figure inclusively. It 
was derived from the pretended quadrature of a M. Van Eick, which 
first gave rise to the discovery. 

Bat the first who ascertained this ratio to any great degree of ex- 
actness was Van Ceulen, a Dutchman, in his book, De Circulo et Ad~ 
scriplis. He found that if the diameter of a circle was 1, the circum- 
ference would be3.141592653589793238462G43383279502884 nearly; 
which is exactly true to 36 places of decimals, and wsls effected by 
the continual bisection of an arc of a circle, a method so extremely 
F 2 



66 MENSUKATION 



EXAMPLES. 



1. If the diameter of a circle be 17, what is the circum- 
ference ? 

Here 3.1416 x 17=53.4072=circMwifererace. 

2. If the circumference of a circle be 354, what is the 
diameter? 

TT 354.000 ,,oco7 J- 
Here =11 2.681 =riza/ne^er. 

3.1416 

3. What is the circumference of a circle whose diame- 
.ter is 40 feet ? Ans. 125.6640. 

4. What is the circumference of a circle whose diame- 
ter is 12 feet? Ans. 37.6992. 

.5. If the circumference of the earth be 25000 miles, 
what is its diameter ? Ans. 7958 nearly. 

6. The base of a cone is a circle ; what is its diameter 
when the circumference is 54 ieQi'l Ans. 20.3718. 



troublesome and laborious that it must have cost him incredible pains. 
It is said to have been thought so curious a performance, that the 
numbers were cut on his tomb-stone in St. Peter's Church-yard, at 
Levden. This last number has since been confirmed and extended to 
double the number of places, by the late ingenious Mr. Abraham 
Sharp, of Little Horton, near Bedford, in Yorkshire. 

But since the invention of Fluxions, and the Summation of Infinite 
Series, there have been several methods discovered for doing the same 
tiling vidth much more ease and expedition. The late Mr. John 
Machin, Professor of Astronomy in Gresham College, has by these 
means given a quadrature of the circle which is true to 100 places of 
decimals; and M. de Lagny, M. Euler, &c, have carried it still 
further. All of which proportions are so extremely near the truth, 
that, except the rrsMo could be completely obtained, we need not wish 
for a greater degree of accuracy. 



OF SITPEHFICIES- 67 

PROBLEM X. 

To find ike length of any arc of a circle* 

RULE * 

1. When the chord of the arc and the versed sine of half 
the arc are given. 

To 15 times the square of the chord, add 33 times the 
square of the versed siiie,t and reserve the number. 

To the square of the chord, add 4 times the square of 
the versed sine, and the square root of the sum vdll be 
twice the chord of half the arc. 

Multiply twice the chord of half the arc by 10 times the 

V * Demon, Put c=:^ the chord of the arc, and i?=:the 
versed sine of half the arc, then the rule may be expressed 
thus : 

, ' V(4c2+«2).10l>2 



;)=2V(— ^-4 



('^60^^ 27.) = 



60c3+33d2^ 

2N/c?«(H-g0^_2^^) = 2v/d^^l-fg5+40^3 +800(^3) 

Now 2Vc^Ml+g^+^H-lfg7+ &c.) is known 

to be the length of an arc whose diameter is d, and the 
versed sine of half the arc v ; and this differs from the 

preceding only by — — -;-> <fec. 
^ s J J' 5600^^3 



+ Here, as in many other places in the following part of the work, 
the term versed sine is used instead of versed sine of half the arc ; 
but in all cases of the kind, it is the versed sine of half the arc that is 
to be understood. 



68 MENSURATION- 

square of the versed sine, divide the product by the re- 
served number, and add the quotient to twice the chord of 
half the arc : the sum will 'be the length of the arc very 
nearly. 

T\^en the chord of the arc, and the chord of half the arc 
are given. — From the square of the chord of half the arc 
subtract the square of half the chord of the arc, the re- 
mainder will be the square of the versed sine : then pro- 
ceed as above. 

2. When the diameter and the versed sine of half the arc 
are given. 

From 60 times the diameter subtract 27 times the versed 
sine, and reserve the number. 

Multiply the diameter by the versed sine, and the square 
root of the product will be the chord of half the arc. 

Multiply twice the chord of half the arc by 10 times the 
versed sine, divide the product by the reserved number, 
and add the quotient to twice the chord of half the arc ; 
the sum will be the length of the arc very nearly. 

Note 1. — When the diameter and chord of the arc are 
given, the versed sine may be found thus : From the square 
of the diameter subtract the square of the chord, and ex- 
tract the square root of the remainder. Subtract this root 
from the diameter, and half the remainder will give the 
versed sine of half the arc. 

2. The square of the chord of half the arc being divided 
by the diameter will give the versed sine, or being divided 
by the versed sine will give the diameter. 

3. The length of the arc may also be found by multiply- 
ing together the number of degrees it contains, the radius 
and th« number .01745329. 

Or, as 180 is to the number of degrees in the arc, so is 
3.1416 times the radius, to the length of the arc. 



i 



OF SUPERFICIES. 69 

Or, as 3 is to the number of degrees in the arc, so is 
.05236 times the radius, to the length of the arc.* 



EXA3IPLES. 

1. If the chord DE be 48, and the versed sine CB 18, 
what is the length of the arc 1 Ans. 64.2959. 




Here 48^ x 15=34560 
182X33=10692 



45252 reserved number. 
48^ =i2S04:=the square of the chord. 
18^ X 4=1296=4 times the square of the versed sine. 

y/S600 =60 =twice the chord of half the arc ACB. 
60X182X10 194400 



Now- 



-=:4.2959, which added to 



45252 ~ 45252 
twice the chord of half the arc gives 64.2959= fAe length 
Gf the arc. 

2. Given the diameter CE 50, and the versed sine CD 
18, what is the length of the arc 1 Ans. 64.2959. 



* When very great accuracy is required, the foUovring' theorem 
may be used. Let d denote the diameter of the circle, and v the 
versed sine of half the arc, then the are = 2 v'' <^» X (1 -|- 

3l52 5v^ 35 y 4 03^5 

— + — + + -f— 

Qd AOd^ ll2d^ 1152cZ4 2816^5 



-+,&c.) 



70 MENSURATION 

50X60=3000 
18X27= 486 



2514 reserved number* 



AC= v'50x 18=30=^^e chord of half the arc. 

80x2x18x10 10800 , ^^_ ,., ,,, 

=— TTTrTT" =4.29o9, wkick added to ttvic 

2514 2514 

the chord of half the arc gives 64.2959=^Ae length of the 

arc ACB. 

3. The chord of the whole arc is 7, and the versed sine 
2, what is the length of the arc ? Ans. 8.4343. 

4. The chord of the whole arc is 40, and the versed sine 
15, what is the length of the arc ? Ans. 53.5800. 

5. The chord of the whole arc is 50, and the chord of 
half the arc 27, rec^uired the length of the arc. 

Ans. 55.3720. 

6. Given the diameter of the circle 100, and the versed 
sine 9, required the length of the arc. Ans. 60.9380. 

7. Given the chord of the whole arc 16, and the diame- 
ter of the circle 20, required the length of the arc. 

Ans. 18.5439. 

8. The diameter of the circle is 50, and the chord of 
half the arc 30, what is the length of the arc ? 

Ans. 64.2959. 

9. The chord of half the arc is 25, and the versed sine 
15, required the length of the arc. Ans. 53,5800. 

PROBLEM XI. 

To find the area of a circle. 

RULE I.* 

Multiply half the circumference by half the diameter, 
and the product will be the area. 

* Demon. A circle may be considered as a regular polygon of an 
infinite nmnber of sides, tJie circumference being equal to the perime- 
ter, and the radius to tlie perpendicular. But the area of a regxilar 
polygon is equal to hall" the perimeter multiplied by the perpendicular. 



OP SUPERPICIES. «1 

Or take i of the product of the whole circumference 
and diameter. 



1. What is the area of a circle whose diameter is 42, 
and circumference 131.946 ? 
2)131.946 



65.973=^ circurnference* 
21 = 5 diarneter. 



1385.433=arm reqidred* 

2. What is the area of a circle whose diameter is 10 
feet 6 inches, and circumference 31 feet 6 inches'? 
fe, in. 

15 9=15.75=^ circumference. 

5 3= 5.25=^ diameter. 



7875 
3150 

7875 

82.6875 
12 

, 8.2500 



Ans. 82 feet 8 inches. 



and consequently the area of a circle is equal to half the eircum- 
fcrence multiplied by the radius, or half the diameter. Q. E. D. 

This rule may be otherwise demonstrated by the doctrine of flux- 
ions. 



72 MENSUBATIOW 

3. What is the area of a circle whose diameter is 1, and 
circumference 3.1416? Ans. .7854. 

4. What is the area of a circle whose diameter is 7, and 
circumference 22 ? Ana. 38^. 

RULE II.* 

Multiply the square of the diameter by .7854, and the 
product will be the area ; or, 

Multiply the square of the circumference by .07958, and 
the product will be the area. 

* Demon. All circles are to each other as tlie squares of their 
diameters. (Euc. XII. 2.) 

But the area of a circle whose diameter is 1, is .7854. &c (bj 

.7854, &c. X d^ 
Kule 1.) Therefore 1^ id^ : : .7654, <^c. : = 

.785, &c. X <Z2=area of a circle whose diameter is d. Q. E. D. - 
The followino- proportions are tlioso of Meiius and Archimedes. 
As 452 : 355 : : square of the diameter ; area- 
As 14 : J 1 : : square of the diameter : area. 
If the circumference be given, instead of the diameter, the area 
may be found as follows : 

The square of the circumference X .07958 =ai'ea. 
As 88 : 7 : : square of the circumference : area. 
As 1420 : 113 : : square of the circumference : area. 
And if cZ be tlie diameter, c the circumference, a the area, and p 
=3.14159, &c. then : 

c 4a a 

p c p 

4a. 

2. c=pd= — =2y/pa 

d 
pd^ C' dc 

3. a=— =— =— 

4 '42> 4 
The following table will also show most of the useful problems 
relating to the circle and its equal or inscribed square. 

1. diameter X .8862=side of an equal square. 

2. circumf. X .2821 = side of an equal square. 

3. diameterX .7071 = side of the inscribed square. 

4. circumf. X .2251=side of the inscribed square. 



OF SUPERFICIES. 73 



EXAIUPLES. 



1. What is the area of a circle whose diameter is 5 1 
7854 

25= sg'ware of the diameter. 



39270 
15708 



19.6350=i^e answer. 

2. What is the area of a circle whose diameter is 7 ? 

Ans. 38.4846. 

3. What is the area of a circle whose diameter is 4.5 ? 

Ans. 15.9043. 

4. How many square yards are there in a circle whose 
diameter is 3^ feet? Acs. 1.0690. 

5. How many square feet are there in a circle whose 
circumference is 10.9956 yards? Ans. 86.5933. 

6. How many square perches are there in a circle whose 
circumference is 7 miles ? Ans. 399300.6080. 

PROBLEM XII. 

To find the area of a sector^ or that part of a circle which 
is bounded by any two radii and their included arc. 

RULE.* 

Find the length of the arc by Problem X. then multiply 
the radius, or half the diameter, by the length of the arc 
of the sector, and half the product will be the area. 

5. areaX .6366=side of the inscribed square. 

6. side of a square X 1.4142=diam. of its circums. circle. 

7. sideof a square X 4.443 =circumf. of its circums. circle. 

8. sideof a square X 1.1 28= diameter of an equal circle. 

9. side of a square X 3.545 =circumf. of an equal circle. 
* The rule for finding the area of the sector, is evidently the same 

as that for finding the area of the whole circle. 

G 



74 MENSURATION 

Note. — If the diameter or radius is not given, add the 
square of half the chord of the arc, to the square of the 
versed sine of half the arc ; this sum being divided by the 
versed sine, will give the diameter. 



1. The radius AB is 40, and the chord BC of the whole 
arc 50, required the area of the sector. 



!l = 8.7750 ==^Ae versed sine of half the arc. 



80x 60 — 8.7750 X 27 = 4563.0760= #Ae reserved number. 



2 X y 8.7750 X e0=52.9906=twic€the chord of half the arc. 

52.9906X8.7750X10 ^, ^^ , . ^ ,, , 

.r.^^ r^^^r. =1.0190 wkick added to twice the 

4563.0750 

chord of half the arc gives 54.0096 the length of the arc. 

, ^ 54.0096X40 _„^ _^ , , 

And =1080.1 920= area of the sector re- 

2 ^ 

quired. 

2. Find the area of the sector, the chord of whose arc 
is 40, and the versed sine of half the arc 15. 

Ans. 558.1250. 

3. Required the area of the sector, the chord of half the 
arc being 30, and the diameter of the circle 100. 

Ans. 1523.4500. 



OF SITPERFICIES. 75 

4. Given the diameter of the circle 50, and the versed 
sine 18, to find the area of the sector. Ans. 803.69875. 

RULE II.* 



As 360 is to the degrees in the arc of a sector, so is the 
area of the whole circle, whose radius is equal to that of 
the sector, to the area of the sector required. 

Note. — For a semicircle, a quadrant^ &c^ take one half, 
one quarter, &;c. of the whole area. 



EXAMPLES. 

1. The radius of a sector of a circle is 20, and the de- 
grees in its arc 22 ; what is the area of the sector ? 

Here the diameter is 40. 

Hence, by Rule II. Prob. XII. the area of the circle= 
402 X .7854= 1600 X .7854=1256.64. 

Now, 360° : 22° : : 1256,64 : 76.7947=ar£a of the sec- 
tor. 



* Demo?u Let r= radius, ^= number of degrees in the 
arc of the sector, and A=its area. 

Then will 4r2 x .78o4=r2 x 3.1416=area of the whole 
circle, and 2rX 3.1416=its circumference. 

Also 360 : 2rX 3.1416 : : d :i^2i?;llH^ length of 
360 

_, 2arX 3.1416 " ^fr^ X 3.1416 

the arc of the sector. But — —- X i X r= — ~— r 

360 360 

=A, by the last rule. And consequently 360 :d::r^X 

3.1416: A. ' Q. E.D. 



76 MENSURATION 

2. Required the area of a sector whose radius is 26, and 
the length of its arc 147 degrees 29 minutes. 

Ans. 804.3987. 

3. Required the area of a semicircle whose radius is 13. 

Ans. 265.4652. 

4. Required the area of a quadrant whose radius is 21. 

Ans. 346.3614. 



PROBLEM Xni. 
Tojlnd the area of a segment of a circle* 

RULE L* 

1. Find the area of the sector, having the same arc with 
the segment, by the last problem. 

2. Find the area of the triangle formed by the chord of 
the segment, and the radii of the sector. 

3. Then the sum, or difference, of these areas, according 
as the segment is greater or less than a semicircle, will be 
the area required. 

Note, — The difference between the versed sine and ra- 
dius, multiplied by half the chord of the arc, will give the 
area of the triangle. 



1. The radius OB is 10, and the chord AC 10; what is 
the area of the segment ABC ? 



* This rule is too evident to need a demonstration. 



4 



or SUPERFICIES. 77 



■^^^^^ 



^^ AC2 100 

CE ~^?r~^~^^^ »er5e<^ dne of half the arc. 



20x 60 — 5x 2 7=1065=^Ae reserved number. 

10x2x5x10 

fnfi5 = .9390, a?2<i this added to twice the chord 

of half the arc gives 20.9390 =t^e length of the arc. 
20.9390 X 10 
—^ — 2 =104,6950= area of the sector OACB. 

OD=OC=CD=5 the perpendicular height of the tri~ 
angle. 

AD=: VA02— OD2= ^75=:8,6603=* the chord of the 
arc. 

8.6603 X 5= 43.301 5= f^e area of the triangle AOB. 
~ 104.6950— 43.3015=61.3935=area of the segment re- 
quired ; it being in this case less than a semicircle. 

2. Required the area of a segment whose height is 2, 
and chord 20. Ans. 26.8783. 

3. Required the area of a segment of a circle, the ra- 
dius being 10, and the chord of the arc 12. Ans. 16.3500. 

4. Required the area of a segment of a circle, whose 
chord is 16, and the diameter of the circle 20. 

Ans. 44.7195. 

5. What is the area of a segment whose arc is a quad- 
rant, the diameter of the circle being 18 feet ? 

Ans. 23.1174. 
G2 



73 MENSUBATIOPf 

6.* What is the area of a segment, who^e arc contains 
300 degrees, the diameter being 50 7 Ans. 1906.8831. 

RULE n.f 

1. Divide the height, or versed sine, by the diameter, 
and find the quotient in the table of versed sines. 

2. Multiply the number on the right hand of the versed 
sine by the square of the diameter, and the product will be 
the area. 

Note 1. — ^When the diameter or versed sine is not given, 
it may be found by the notes, page 68 or 74. 

Note 2. — ^When the quotient arising from the versed 
sine divided by the diameter, has a remainder or fraction 
after the third place of decimals ; having taken the area 



* The chord of the arc will evidently be equal to the radius of the 
circle. 

+ The table to which this rule refers, is formed of the areas of the 
sf'grnents of a circle whose diameter is 1 ; and which is supposed to be 
divided by perpendicular chords into 1000 equal parts. 

The reason of the rule itself depends upon this property — That the 
versed sines of similar seg-ments are as the diameters of the circles to 
wliich they belong-, and the area of those segments as the squares of 
the diameters ; which may be thus demonstrated. 

Let ADBA and adha be any two similar segments, cut off from the 
similar sectors adbca and adbca, by the chords ab and ab, and let the 
perpendicuJar co bisect them. 

Tlien by similar triangles, dr: 
db :: ^c : be and its : db : : vm : dn ; 
whence, by equality, k: : fee : : mn : dn, 
or 2bc : 26c : : i>m : dn. 

Again, since similar segments are as 
the squares of their chords, it will be 
AB- : alfi :: seg. adba : seg. adba ; but 
AB- : a&- : : cb- : c&-, whence, by equal- 
ity, seg. adba : seg. adba : : cu- : efe2, 
or seg. adba : seg. adia : : 4cb- : 4cfe2, Q. E. D. C 

Now, If d be pat equal to any diameter, and v the versed sine, it 

V }■■' 

will be d:vt:l (diameter in the table): — = versed sine of a similar 

d 
segment in the table whose area let be called a. 




OF SUPERFICIES. 79 

answering to the first three figures, subtract it from the 
next following area, multiply the remainder by the said 
fraction, and add the product to the first area, then the 
sum will be the area for the whole quotient. 



1. K the chord of a circular segment be 40, its versed 
sine 10, and the diameter of the circle 50, what is the 
area? 



5.0)1.0 



.2=tahular versed sine. 
A1182S =ztabular segment. 
2500= square of 50. 



55911500 
223646 



279.557500=area required. 

2. The chord of the segment is 20, the versed sine 5, 
what is the area ? Ans. 69.889375. 

3. The diameter of a circle is 40, and the versed sine 
10 ; what is the area of the segment ? Ans. 245.6736. 

4. If the diameter be 52, and the versed sine 2, what is 
the area of the segment? Ans. 26.9197. 

5. If the chord of half the arc be 80, and the versed sine 
9, what is the area of the segment 1 Ans. 350.1100. 

6. The diameter of a circle is 100, and the chord of the 
arc 60 : what is the area of the segment 1 

Ans. 408.75. 



Then 1'^: d^ :: a : ad^ =area of the segment whose height is v, and 
diameter (2 as in the rule. 



80 WtENSURATIOX 

PROBLExM XIV. 

To find the area of a circular zone, or the space included 
between any two parallel chords and their intercepted 
arcs, 

RULE.* • 

From the greater chord subtract half the difference be- 
tween the two, multiply the remainder by the said half dif- 
ference, divide the product by the breadth of the zone, and 
add the quotient to the breadth. To the square of this 
number add the square of the less chord, and the square 
root of the sum will be the diameter of the circle. 

Now, having the diameter EG, and the two chords AB 
and DC, find, by Prob. XIII. Rule II. the areas of the seg- 
ments ABE A, and DCED, the difference of which will be 
the area of the zone required. 

Note 1. — The difference of the tabular segments multi- 
plied by the square of the circle's diameter will give the 
area of the zone. 

Note 2. — When the larger segment AEB is greater than 
a semicircle, find the areas of the segments AGB, and 
DCE, and subtract their sum from the area of the whole 
circle, the remainder will be the area of the zone. 



* The reason of this rule is too obvious to require a demonstration. 

Note. — When the two parallel sides of the zone are equal, the chord 
of the small segment will be equal to the breadth of the zone, and the 
height of this segment will be equal to >/:^ab--|- ias2=^AB ; as being 
put for the breadth of the zone. 

And when one of the sides is the diameter of the circle, the chord 
of the ^me segment will be \/ as^J^xfiy and its height=?iAB 
— \/|ab2 — ias2 — Id'J where 0=|ab — dc. 



OF SUPERFICIIJS. 81 



1. The greater chord AB is 20, the less DC 15, and 
*heir distance Dr 17^ : required the area of the zone 
ABCD. Ans. 395.4388. 




80—15 

— - — =2.5=^ the difference between the two chords* 

(20— 2.5) X 2.5 
17.5+ ^ Y^ =17.5 + 2.5=20=DF. 

And >/202 -f 152 = ^625=25=fAe diarneter of the cir. 
The segment AEB being greater than a semicircle, we 

^nd by note 1, page 68, the versed sine of DCE=2.5, and 

ihat of AGB=5. 

2.5 
Hence by Prob. XIIL Rule IL, —-= .100 = tabular 

versed sine of DEC. 
5 
And ■^=.20Q=^tabular versed sine of AGB. 

Nmo .040875 X25=area of seg. DEC=25.546875 
And .111823 X 252 =area of se^. AGB=69.889375 

sum 95.43625 
By Prob, XL Rule 11, .7854 X 25^ > ^490.37500 

=zarea of the whole circle, ^ ' 

Diference=area of the zone ABCD= 395.43875 



82 MENSIT»ATIO:^ 

2. The greater chord is 96, the lesser 60, and the 
breadth 26 j what is the area of the zone ? 

Ana. 2136.7500. 

3. One end of a circular zone is 48, the other end is 30, 
and the breadth is 13 j what is the area of the zone ? 

Ans. 534.1875. 

PROBLEM XV. 

To find the area of a circular ring, or the space included 
between the circumference of two concentric circles, 

RULE.* 

The difference between the areas of the two circles will 
be the area of the ring. 

Or, multiply the sum of diameters by their difference, 
and this product again by .7854, and it will give the area 
required. 

EXAMPLES. 

1. The diameters AB and CD are 20 and 15 ; required 
the area of the circular ring, or the space included between 
the circumferences of those circles. 



"^ Demon, The area of the circle aeba=ab2 x .7854, 
and the area of the small circle cd is=CD2 x .7854 ; there- 
fore the area of the ring^AB^ x .7854 — cd^ x .7854= 
AB + CD X AB CD X .7854. Q. E. D. 

Coroll. If CE be a perpendicular at the point c, tlie area of the ring 
will be equal to that of a circle whose radius is CE. 

Rule 2. Multiply half the sum of the circumferences by half tlie 
difference of the diameters, and the product will be the area- 

Tliis rule will also serve for any part of the ring, using half the 
earn of the intercepted arcs for half the sum of the circumfercr^ces. 



OP StJPERPICIES. 83 

Here AB + CD x AB— CD=: 35 x 5= 175 ; and 175 X 

.7854= 137. 4450= area, of the ring required, 

2. The diameters of two concentric circles are 16 and 
10 ; what is the area of the ring formed by those circles? 

Ans. 122.5224. 

3. The two diameters are 21.75 and 9.5, required the 
area of the circular ring. Ans. 300.6609. 

4. Required the area of the ring, the diameters of whose 
bounding circles are 6 and 4. Ans. 15.708. 

PROBI^M XVI. 

Tojind the areas of limes, or the spaces beticeen the inter- 
secting arcs of two eccentric circles, 

RULE.* 

Find the areas of the two segments from which the lune 
is formed, and their difference will be the area required. 



* Whoever wishes to be acquainted with the properties of Junes, 
and the various theorems arising- from them, may consult Mr. Wis- 
tori's Cojnmentary on Tacquefs Euclid, where they will find this sub- 
ject very ing-enioiisly- managed. 
The following property is one of the most curious : 
If ABC be a right angled triangle ; and semicircles be described on 
the three sides as diameters, then will the said triangle be equal to the 
two limes D and F taken together. 




A B 

For the semicircles described on AC and BC=:the one described on 
AB (31.66,) from each take the segments cut off by AC and BC, then 
will the lunes AFCE and BDCG=the triangle ACB. Q. E. D. 



84 



MENSrHATION 



The length of the chord AB is 40, the height DC 10, 
and DE 4 ; required the area of the lune ACBEA. 




By note, page 74, the diameter of the circle of which 

202 -^102 

ACB is a part= — =50. 

And the diameter of the circle of which AEB is a part 

=?21±11=104. 
4 

Now having the diameter and versed sines, we Jind by 

Prob, XIII. Rule III. 

The area of seg. ACB=.111823X 502=279.5575 

jLn<Z area of 5e«^. AEB= .009955 X 1042 = 107.6733 



Their difference is the area of the 'i _^'^-, qqao 
lune AEBCA, required^ \ ~ 

2. The chord is 20, and the heights of the segment 10 
and 2 ; required the area of the lune. Ans. 130.287. 

3. The length of the chord is 48, and the heights of the 
segments 18 and 7 ; required the area of the lune. 

Ans. 408.7057. 



PROBLEM XVII. 

To find the area of an irregular polygon, or a figure cf 
any number of sides. 



OF SUPEHPICIES. 85 



RULE,* 



1. Divide the figure into triangles and trapeziums, and 
find the area of each separately. 

2. Add these areas together, and the sum will be equal 
to the area of the whole polygon. 



EXAMPLES. 

1. Required the area of the irregular figure ABCDEFGA, 
the following lines being given. 

GB=30.5 A7i=11.2, CO=6 
GD=:29 F^^ll Cs=6.6 
FD=:24.8 Ep=4 




F^- 



gH 



* When any part of the figure is bounded by a carve, the area may 
be found as follows : 

Rule 1, Erect any number of perpendiculars upon the base, at equal 
distances, and find their lengths, 

2. Add the lengths of the perpendiculars, thus found, together, and 
the sum divided by their number will give the mean breadth. 

3. Multiply the mean breadth by the length of the Imse, and it 'wiH 
give tlie area of that part of the figure required. 

H 



J 



gg JTEJfsrRATION 

Here 4^±^ x GB= ^^•^ + ^ x 30.5 = 8.6x 30.5 = 
2 2 

262.3=area of the trapezium ABCG. 

An(;?lil2f X GD=li±^^K 29 = 8.8X29=255,2 
2 2 

rrzarea of the trapezium GCDF. 

,j FDxEw 24.8X4 99.2 .^ ^ ^ ,, ,. 

Also^ Lz=i = = 49.6= area of the tn- 

2 2 2 *" 

an^Ze FDE. 

Whence 262.3 + 255.2 + 49.6= 567.1 =area of the who. 
figure required. 

2. *In a pentangular field, beginning with the south side 
and measuring round towards the east, the first, or south 
side, is 2735 links, the second 3115, the third 2370, the 
fourth 2925, and the fifth 2220 ; also the diagonal from the 
first angle to the third is 3800 links, and that from the third 
to the fifth 4010; required the area of the field. 

Ans. in acJi ro, 29 po. 

Promiscuous Questions concerning Lines and Areas. 

1. Given AC=32, AD=3, EC =8, the perpendicular 
DG=4, and the perpendicular EF=6, to find the area of 
the triangle ABC and the sides AB and BC. 



To find the area of mixed or compound figures, or such as are com- 
posed of rectilineal and curvilineal fio-ures together ; the rule is to find 
the area of the several figTires of which the whole figure is composed, 
then add all the areas together, and the sum will be the area of the 
whole, compound figure. And in the same manner may any irregular 
field or piece of land be measured, by dividing it into trapeziums and 
triangles, and finding the area of each separately, 

* Note. As this figure consists of three triangles, all of whose sides 
are given, by calculatinjj- their areas according to Problem III. the 
sum will be the area of the whole figure accurately, without drawing 
perpendiculars from the angles to the diagonals. 

The same thing may also be done in most other cases of this kind. 




m 



A D H I 



Draw GH parallel to BC, then since the sides about the 
equal angles in equiangular or similar triangles are propor- 
tional, we have in the similar triangles CEF and HDG. 
FE : EC :: GD : DH=VS which added to AD will give 
AH=V- Also in the sim. A's AGH, ABC, AH : AC 
: : GD : 61=15.36, and this multiplied by half the base will 
give the area of ABC=245.76. 

Again in each of the right angled triangles ADG and 
CEF we have the two legs to find the hypothenuses AG= 
6 and CF=10. Now by sim. a's ADG, AIB, GD ; GA 
: : BI : BA=19.2 ; and FE : FC : : BI : BC. 

2. K from the right angled triangle ABC, whose base is 
12, and perpendicular 16 feet, be cut off, by a line DE par- 
allel to the perpendicular, a triangle whose area is 24 square 
feet: what are the sides of this triangle? 



B/ 



A D B 

The area of the triangle ABC=ABxiBC=96, also 
iaving AB and BC, AC m.ay be found =20. 

Now it is evident that the triangles ABC and ADE are 
similar, and since the areas of sim. a's are as the squares 
of their like sides, we have. 



88 KENSUHATION 

Area ABC : area ADE : : AC^ : AE 
Area ABC : area ADE : : BC^ : DE» 
Area ABC : area ADE : : AB^ : AD^ 
From which AE=10, AB=6, and DE=8. Ans. 

3. A gentleman in his yard has a circular grass-plot, the 
diameter of which is 25 yards. Query the length of the 
string that would describe a circle to contain nine times aa 
much. Ans. 37.5 yards, 

4. Suppose a ladder 100 feet long, placed against a per- 
pendicular wall 100 feet high, how far would the top of the 
ladder move down the wall by pulling out the bottom there- 
of 10 feet? Ans. .5012563. 

5. *There is a circular pond whose area is 5028^ square 
feet, in the middle of which stood a pole 100 feet high : 
now the pole having been broken, it was observed that the 
top just struck the brink of the pond ; what is the height 
of the pole? Ans. 41.9968. 

6. fin a level garden there are two lofty firs, having 
their tops ornamented with gilt balls; one is 100 feet high, 
the other 80, and they are 120 feet distant at the bottom ; 
now the owner wants to place a fountain in a right line be- 
tween the trees, to be equally distant from the top of each; 
what will be its distance from the bottom of each tree, and 
also from each of the balls ? 

C From the bottom of the lower tree 75 feet. 
Ans. < From the bottom of the higher tree ^5 feet, 
( From each ball 109.6585/ee^ 



* This problem may be constructed by forming a right angled tri- 
angle, having the radius of the circle for the base, and the length of 
the pole for the perpendicular ; and erecting a perpendicular on the 
middle of the hypothenuse to cut the perpendicular of the triangle ; 
this will determine the place where the pole was broken. 

t The figure to this question is thus constructed. Draw AC=120, 
the distance of the trees at the bottom, and erect the perpendiculars 



OF SUPERFICIES. 89 

7. A person wishes to inclose 6ac. Iro. l2po, in a tri- 
angle similar to a small triangle whose sides are 9, 8, and 6 
perches respectively ; required the sides of the triangle. 

Ans. 59.029, 52.47, and 39.353 perches, 

8. Required the sides of an isosceles triangle, containing 
6ac. Oro. \2per* and whose base is 72 perches. 

Ans. 45 perches each. 



AE= height of the lower tree, and CI>=tiie higher. Join ED, and 
from the middle of it draw the perpendicular GF, and F will represent 
the place of the fbimtaiii. Join EF and DF, and draw EI parallel to 
AC, and GB parallel to DC ; then the triangles EID and GBF beings 

eimilar, the calciilation is evident 




112 



CONIC SECTIONS. 



DEFINITIONS. 



1. The conic sections are such plain figures as are fonned 
by the cutting of a cone. 

2. *A cone is a solid described by the revolution of i 
right-angled triangle about one of its legs, which remains 
fixed. 



3. The axis of the cone is the right line about which the 
triangle revolves. 



* This is Euclid's definition of a cone, and is that which is gene- 
rally best vmderstood by a learner ; but the following- one is more 
general. 

Conceive the right line CB to move upon the fixed point C as a 
centre, and so as continually to touch the circumference of the circle 
AB, placed in any position, except in that of a plane which passes 
through the said point ; and then that part of the Une which is inter- 
cepted between the fixed point and the periphery of the circle will 
generate the convex superficies of a cone. 



CONIC SECTIONS. 91 

4. The hose of a cone is the circle which is described by 
the revolving leg of the triangle. 

5. If a cone be cut through the vertex, by a plane which 
also cuts the base, the section will be a triangle. 




6. If a cone be cut into two parts, by a plane parallel to 
the base, the section will be a circle. 




7. If a cone be cut by a plane which passes through its 
two slant sides in an oblique direction, the section will be 
an ellipsis* 




8. The longest straight line that can be drawn in an el- 
ipsis is called the transverse axis ; and a line drawn per- 
pendicular to the transverse axis, passing through the centre 
of the ellipse, and terminated both ways by the circumfer* 
ence, is called the conjugate axis^ 



92 



CONIC SECTIONS. 




9. An ordinate is a right line drawn from any point of 
the curve, perpendicular to either of the diameters. 




10. An abscissa is that part of the diameter which is 
contained between the vertex and the ordinate. 

11. If a cone be cut by a plane, which is parallel to 
either of its slant sides, the section will be a parabola. 




12. The axis of a parabola is a right line drawn from 
the vertex, so as to divide the figure into two equal parts. 




13. The ordinate is a right line drawn from any point in 
the curve perpendicular to the axis. 

14. The abscissa is that part of the axis which is con- 
tained between the vertex and the ordinate. 



CONIC SECTIONS. 



9S 



15. *If a cone be cut into two parts, by a plane, which, 
being continued, would meet the opposite cone, the section 
is called an hyperbola* 




16. The transverse diameter, or axis of an hyperbola, is 
that part of the axis intercepted between the two opposite 
cones. 

17. The conjugate diameter is a line drawn through the 
centre perpendicular to the transverse. 

18. An ordinate is a line drawn from any point in the 
curve perpendicular to the axis; and the abscissa is the 
distance intercepted between that ordinate and the vertex. 

PROBLEM I. 

To describe an ellipsis, the transverse and conjugate diart^ 
eters being given. 




* The two opposite cones, in this definition, are supposed to be gen- 
erated together, by the revolution of the same line. 

All the figures which can possibly be formed by the cutting of a 
cone, are mentioned in these definitions, and are the five following 



94 CONIC SECTIONS* 

Construction.* 1. Draw ihe transverse and conjugal© 
diameters, AB, CD, bisecting each other perpendicularly in 
the centre o. 

2. With the radius Ao, and centre C, describe an arc 
cutting AB in Ff; and these two points will be the foci of 
the ellipse. 

3. Take any number of points nn, &c. in the transverse 
diameter AB, and with the radii Aw, wB, and centres F^, 
describe arcs intersecting each other in 5, s, &c. 

4. Through the points 5, s, &c. draw the curve AsCBD, 
and it will be the circumference of the ellipse required. 



PROBLEM II. 

In an ellipsis^ any three of the four following terms being 
given, viz, the transverse and conjugate diameter s.^ an 
ordinate and its abscissa, to fnd the fourth, 

CASE I. 

When the transverse, conjugate, and abscissa are known, to 
fnd the ordinate. 



ones : tiz. a triangle, a circle, an ellipsis, a parabola, and an hyper- 
bola ; but the last three only are usually called the conic sections. 

* It is a known property of the ellipse, that the sum of two lines 
drawn from the foci, to meet in any point in the curv'e, is equal to the 
transverse diameter, and from this the truth of the construction is 
e^^dent. | 

From the same principle is also derived the following method of 
describing an ellipse, by means of a string and two pins. 

Having found the foci F,/, as before, take a tlircad of the length of 
the transverse diameter, and fasten iis ends with two pins in the points 
F,/; then stretch the thread F sf to its greatest extent, and it will 
reach to the point s in the curv^e ; and by moving a pencil round within 
the thread, keeping it always stretched, it will trace out the curve re- 
quired. 



coiac SECTIONS. 95 

RULE.* 

As the transverse diameter is to the conjugate, 
So is tlie square root of the rectangle of the two ab- 
scissas, 

To the ordinate ^vhich divides them. 

EXA3IPLES. 

1. In the ellipsis ADBC, the transverse diameter AB ig 
120, the conjugate diameter CD is 40, and the abscissa BF 
24 ; what is the length of the ordinate EF ? 

C E 




Here 120 (AB) : 40 (CD) :: V 96x24 (AFxFB) i 

40 

J2Q \/96x24=ix/2304=iX48=16 = EF the ordinate 

required^ 

2. If the transverse diameter be 35, the conjugate 25, 
and the abscissa 28 ; what is the ordinate? Ans. 10. 

CASE U. 

When the transT>erse, conjugate, and ordinate are known^ 
■ to find the abscissa^ 



* Let f=the transverse diameter, c=conjugate, a:=any 
abscissa, and 2/= ordinate. Then will the general equation 
expressing the property of the ellipse, hefit^iixX {t — x) 
I y^ ; and from this the four rules here given are deduced, 



the one above being y=—-y/xXt 



96 CONIC SECTIONS. 

RULE.* 

As the conjugate diameter is to the transverse, 

So is the square root of the difference of the squares of 
the ordinate and semi-conjugate, 

To the distance between the ordinate and centre. 

And this distance being added to and subtracted from 
the semi-transverse, will give the two abscissas required. 



1. The transverse diameter AB is 120, the conjugate di- 
ameter CD is 40, and the ordinate FE is 16 ; what is the 
abscissa FB 1 



Here 40(CD) : 120(AB) :: v/202— 162(VOB2~FE2) : 

120 

— V202—16-=3V400—256=3v/144=3x 12=36 = 

OF, the distance from the centre. 

Whence 60(OB)— 36(OF) = 24=BF > _ . , . 

A7icZ60(OA) + 36(OF)=96=AF5 - ^"''^ abscissas 
required. 

2. What are the two abscissas to the ordinate 10, the 
diameters being 35 and 25 ? Ans. 7 and 28. 

CASE III. 

When the conjugate^ ordinate^ and abscissa are knowriy to 
Jind the transverse, 

RULE.t 

1. To or from the semi-conjugate, according as the 
greater or less abscissa is used, add, or subtract the square 



* This rule in algebraic terms is as follows : The greater abscissa 

t t t t 

ar= — \-- 's/\c^ — y- or the less ar= Vic- — y^. 

2 c 2 c 

t This rule in algebraic terms is as follows: t={c+2 



coTfic sEcnojfs. 97 

root of the difference of the squares of the ordinate and 
semi -conjugate. 

2. Then, as this sum or difference, is to the abscissa, 

So is the conjugate to the transverse. 



EXAMPLES, 

1. The conjugate diameter CD is 40, the ordinate EF 
is 16, and the abscissa FB 24 ; required the transverse 
AB. . 

Here 20— v^ 20^— 162(^002— EF2)r=:20— 12=8. 
And 8 ; 24 : : 40 : 120, the transverse diameter required. 

2. If an ordinate and its lesser abscissa be 10 and 7, and 
the conjugate 25, what is the transverse ? Ans. 35. 

CASE IV. 

The transverse^ ordinate, and abscissa being given, to find 
the conjugate. 

RULE.* 

As the square root of the product of the two abscissas, 
is to the ordinate, so is the transverse diameter, to the con- 
jugate. 

EXAMPLES. 

1. The transverse AB is 120, the ordinate EF 16, and 
the abscissa FB 24 : required the conjugate. 



ex 



v/ \c^ — y- ) X -^, or f = (c — -2 V ic^ — y^) X ? according 

as the greater or less abscissa is used. 

*'The rule in algebraic terms is tyX =c, the 

s/ tX XX 

conjugate, or shortest diameter. 
I 



98 CONIC SECTIONS. 

Here v/ 24x96 js/ WxAF) : 16 (EF) : : 120 (A B) 
: 16X120-^V'24X96 = 16X120-^V2304=16X]204- 

,^ 16X120 120 ,^ , . J- , -7 

48 r= = =40 the conjugate diameter required. 

48 3 

2. The transverse diameter is 35, the ordinate is 10, and 

its abscissa 6 : what is the conjugate ? Ans. 25. 

PROBLEM III, 

Tojind the circumference of an ellipse^ the transverse arid 
conjugate diameters being known, 

RULE.* 

Multiply the square root of half the sum of the squares 
of the two diameters by 3.1416, and the product will be 
the circumference nearhj. 

* Demon. Let <=transverse diameter, c=conjug^te, p=3,1416, 
and d=l-|. Then will p X (_iL__|l_ _ 

' &;c.)= circumference of the ellipse, as is shown 

by the writers on fluxions. 

t'^c^ 1 c^ 

Now the rule given above isp^/ — =:pty/~ + —-^) 

1 c^ d 

t=pt^(l-^--\-~)=(bj substitution) p<v/(l—-)=_pfX 

d d^ Sd^ 

(I ), &,c. But the first tUree terms of this 

^ 2.2 23.4 24.4.6^ 

series differ from the first three terms of the former only by 

— ; therefore the rule is shown to be an approximation. Q. E. D. 
64 

Rule 2. Multiply J the sum of the two diameters by 3.1416, and the 
product will give the circumference exact enough to answer most prac- 
tical purposes. 

Rule 3. Find the circumference both from the last rule and that 
given above, and ^ the sum of tlie results will give the circumference 
extremely near. 



CONIC SECTIONS* 90 



1. The transverse diameter is 24, and the conjugate 20: 
required the circumference of the ellipse. 

AB2 + CD2 24^+202 ,576 + 400 
^ere V -—^ — ^=V ^ =V ^ = 



v/ 288 + 200== >/488= v/ 22.09. 

And 22.09X3.1416^=69.397944 the circumference re- 
quired, 

2. The axes are 24 and 18 : what is the circumference ? 

Ans. 66.6434. 

PROBLEM IV 

To find the area of an ellipse^ the transverse and conjugate 
diameters being given* 

RULE.* 

Multiply the transverse diameter by the conjugate, and 
the product again by .7854, and the result will be the area. 

Or multiply .7854 by one axe, and the product by the 
other. 



Note. — If o = semi-transverse BO, C= semi-conjugate CO, and 
a:= distance OF, of the ordinate EF from the centre, then will the 

arc CE be = a! X ( 1 + g^ x^ + —^^ x^ + 



112^12 



- x^ &c. 



Tlie following- may serve as a practical rule for finding the length 
of the arc CE. 

Find the length of a circular arc intercepted by OE and OC, and 
wliose radius is J the sum of OE and OC, and it will be the elliptie 
arc CE nearly . 

* The demonstration of this rule is contained in that of the nexi 
|iroblem. 



100 CONIC SECTI0PT9. 



EXAMPLES. 



1. What is the area of an ellipse whose transverse di- 
ameter is 24, and the conjugate 18 ] 

Here 24x 18x .7854=339.2928=area required, 

2. If the axes of an ellipse be 35 and 25, Avhat is the 
area ? Ans. 687.225. 

3. Required the area of an ellipsis whose two axes are 
70 and 50. Ans. 2748.9. 

PROBLEM V. 

To find the area of an elliptic segment, whose base t* 
parallel to either of the axes of the ellipse, 

RULE.* 

1. Divide the height of the segment by that axe of the 
ellipse of which it is a part, and find in the table a circular 
segment, whose versed sine is eftjual to the quotient. 



* Demon. Let the transverse diameter 2ab=;, the con- 
jugate CD=c, and AG=ar, and eg=2/J then by the property 



of the curve we shall have y=.~s/tx — x^^ and the fluxion 



of the area EAF=(2/a:)=-X x^Jtx — x-. But xX y/tx — x^ 

is known to express the fluxion of the corresponding circu- 
lar segment, whose versed sine is ar, and the diameter t. 
Let the fluent of this expression therefore be denoted by a, 



and then the fluent of— Xxy/tx — a:- will be = — X a, 
from whence the rule is derived. Q. E. I. 

Corol. The ellipse is equal to a circle whose diameter is a mean 
proportional between the two axes, and from hence the rule is formed 
for the whole ellipse. 



CONIC SECTIONS. 101 

2. Multiply the segment thus found, and the two axes 
of the ellipse continually together, and the product will 
give the area required. 

EXAMPLES. 

1. Required the area of the elliptic segment EAF, whose 
height AG is 10, and the axes of the ellipse 2AB and CD, 
35 and 25 respectively* 



•D 



Here 



10.000 2.0000 



= ,2851 =z tabular versed dm. 



35 7 

And the tabular segment belonging to this is .185153. 
Whence .185153 X 35 (2AB) X 25(CD)=6.480355x 25 
= 162.0088= area of' the segment required. 

2. What is the area of an elliptic segment cut off by a 
double ordinate parallel to the conjugate axe, at the dis- 
tance of 36 from the centre, the axes being 120 and 40? 

Ans. 536.7504. 

3. What is the area of a segment cut off by an ordinate 
parallel to the transverse diameter, whose height is 5, the 
axes beino^ 35 and 25 ? Ans. 97.845125. 



The area of an elliptic segment may also be found by the following 
rule. 

Find tlie corresponduig seg'ment of the circle described upon the 
same axe to which the base of the segment is perpendietdar. 

Then as this axe is to the other axe, so is the circular segment Jo 
ihe elliptic segment. 

12 



102 



COXIC SECTIONS. 



PROBLEM \l. 



To describe a parabola, any ordinate to the axe and its 
abscissa being given. 





c 






-V 






s/ 


F 


X 


-Vj» 


/ 


••'•f. 


....; 


\% 


f 


7> 

R 




\ \ 


S 


-> 


..-■ 


771 S 



71 



* Construction. 1. Bisect the given ordinate RS in m; 
join Vwi, and draw mn perpendicular to it, meeting the axe 
in 72. 

2. Make VC and VF each equal to Rn, and F will be 
the focus of the curve. 

3. Take any number of points r, r, &;c. in the axe, 
through which draw the double ordinates SrS, &;c. of an 
indefinite length. 



* Since Ymn is a right angled triangle, and mR is perpendicular 
to vn, vX ^K,n=VR X VF=Rrn^=^Rs2, which is a known property 
of the parabola when F is the focus. And because SF2=cr2 — rrz 
= cr -{- JF X cr — 7F = cr _[- rF X cf = 2vr X 2vf =. vr X 4vf, 
therefore, s is a point in the curve of a parabola, and the same may be 
shown of any other point s. 

Besides the methods above, for finding the focus, it may be foimd 
- arithmetically as follows : 

Divide the square of the ordinate by 4 times the abscissa, and tlie 
quotient will be the focal distance VF. 

Several other metliods of doing tliis, as well as of describing the 
curve itself, may also be found in Emerson's Conic Sections, and other 
performances. 



CONIC SECTIONS. 103 

4. With the radii CF, Cr, &c. and centre F, describe 
arcs cutting the corresponding ordinates in the points s, s, ■ 
&c. and the curve SVS drawn through all the points of in- 
tersection will be the parabola required. 

Note. — ^.The line sFs passing through the focus F is 
called the parameter. 

PROBLEM Vn. 

In a parabola, any three of the four following terms being 
given, viz. any two ordinates and their two abscissas, to 
fnd the fourth, 

RULE.* 

As any abscissa is to the square of its ordinate, so is any 
other abscissa to the square of its ordinate. 

Or as the square root of any abscissa is to its ordinate, 
so is the square root of any other abscissa to its ordinate. 

EXA3IPLES. 

1. The abscissa VF is 9, and its ordinate EF 6 ; required 
the ordinate GH, whose abscissa VH is 16. 

V 



E 
G 



/■ 






H 



\ 



Here V9(VVF) : 6(EF) : : v/16(v/VH) : ^^fj'^ ■■ 
——=—=^8= ordinate GH. 



* If X and X be any two abscissas, and y and Y their corresponding^ 
ordinates, the equation of the ciirve will be xY'^:=Xy% which is the 
same as the rule. 



104 COXrC SECTIONS. 

Or, 

9(VF) : 36(EF^) : ; 16 (VH) :i.?^^=, iq ^ 4 = 64 = 

Gff, or 8= GH as before, 

2. The two abscissas are 9 and 16, and their correspond- 
ing ordinates 6 and 8 ; from any three of these to find the 
fourth. 

PROBLExM Vin. 

To find the length of any arc of a parabola, cut off by 
a double ordinate, 

RULE.* 

To the square of the ordinate add | of the square of the 
abscissa, and tuice the square root of the sum will be the 
length of the curve required. 



* Demon, Let .t= any abscissa, i/= its ordinate, a= 

y 

\ the pm-ameter of the axe, and ^=— . Then it is shown 



by the writers on fluxions, that, aqy/\-\-q^-\-aX hyp. log. 
(f q^ Sq' 3.5o« 

of (?+ ^/l+^y)=%x(l + ^3-.^4:5 + 2X6^7-.2X6:8:9 

(f 

&c.)=c=length of the curve. But ^ 1 +^^'=1 +^3— 

q' ^' . . 

^ Q + 4 f 21 ^'^^' ^S^^^^^S ^^'i*h the former in the tAvo 
first terms. 

Therefore — = s/l-rlq^ nearly ; and consequently c= 
2y v' 1 + i?'= 2 n/ .vM^^ the same as the rule. ^ Q. E. D. 

Note. — This rule must be used only when the abscissa does not 
exceed half tlie ordinate. The length of the curve in other cases 
must be found by means of hyperbolic logarithms, as is shown by 
writers on fluxions. 



COXIC SECTIONS* 



EXAMPLES. 



105 



1. The abscissa VH is 2, and its ordinate GH 6 ; what 
is the length of the arc GVK 1 

Here 2^(yW) x f -f 26 (GIF) = ^ + 36= V + 36= 

5.333, &c. + 36=41.333333. 
And 41.333333(6.429 
36 2 



124)533 12.858= Zen^A of the arc. 

496 

1282)3733 
2564 



12849)116933 
115641 



1292 



PROBLEM IX. 

To find the area of a parabola, its base and height being 
given. 

RULE.* 

Multiply the base by the height, and f of the product 
will be the area required. 

* Demon. Let vH=a!, gh=2^j and the parameter =p. 

Then px^='if, or ^px=y by the nature of the curve. 

Whence the fluxion of the area (=y^)=^y/px and its 
fluent = fa; X Vpx. 

But because y=y/pxj therefore f xX 's/px=ixy=aiea. 
of the parabola, which is the same as the rule. 

Coroll. Every parabola is =f of its circumscribing' 
parallelogram. 



106 CONIC SECTIONS, 



EXAMPLES. 

1. What is the area of a parabola GVK, whose height 
VH is 12, and the base or double ordinate GK 16 ? 

V 




K 

16 X 12x 2 
Here 16(GK)x 12(VH)X t= 3 = 16X4X2= 

128= area required. 

1. The abscissa is 12, and the double ordinate or base 
38 ; what is the area 1 Ans. 304. 

3. What is the area of a parabola whose abscissa is 10, 
and ordinate 8] Ans. 106f. 

PROBLEM X. 

To find the area of afrustriim of a parabola. 

RULE.* 

Divide the difference of the cubes of the two ends of the 
frustrum by the difference of their squares, and this quotient 
multiplied by | of the altitude, will give the area required. 

* Demon. Let d=gk, fZ=Ei, and a=FH. 

Then by the nature of the curve d^ — cP : a : : d^ : 

AD^ 1 2 J2 J' «^^ 

=VH, and D^ — d^ ; a : : a" : - 



jy'—d^ ' • " 'ji'—d' 

. , ' r 9 00^ - ad^ , d' — rf^ 

And therefore f X ^2_^^2 — I X ^,_^, = § a X ^^^ 

=:area of the frustrum. Q. E. D. 

Note. — Any parabolic frustrum is equal to a parabola of the same 
altitude, whose base is equal to one end of the frustrum, increased 



COTTIC SECnOTJS. 



107 



EXAMPLES. 

1. In the parabolic frustrum GEIK, the two parallel 
ends EI and GK are 6 and 10, and the altitude, or part of 
the abscissa FH, is 3 ; what is the area ? 
V 



EA 



G 



H 



Here 10^— 63(GK' — EP) -^ 10^ — 6^(GK2 — EI^ = 

103_63 ^ 1000_2i6 _78498_49_ ^5 

102_62~ 100— 36 ~'6r~T"~T'~ 

2x3 

And 12.25X =12.25 X 2=24.5 = area required. 

3 

2. The greater end of the frustrum is 24, the lesser end 
is 20, and their distance 5^ j what is the area ? 

Ans. 121.3333, 

3. Required the area of the parabolic frustrum, the 
greater end of which is 10, the less 6, and the height 4. 

Ans. 32|, 

PROBLEM XL 

To construct an hyperbola, the transverse and conjugate 
diameters being given^ 




by a third proportional to the sum of the two ends, and the other 
end. 



108 CX)NIC SECTIONS. 

* Construction, 1. Make AB the trans veree diameter, 
and CD perpendicular to it, the conjugate. 

2. Bisect AB in O, and from O with the radius OC, or 
OD, describe the circle DfCF, cutting AB produced in F 
andy, which points will be the two foci. 

3. In A3 produced take any number of points, w, w, 6lc. 
and from F and/', as centres, with the distances Bra, An, as 
radii, describe arcs cutting each other in 5, 5, &;c. 

4. Through the several points s, 5, &c. draw the curve 
«Bs, and it will be the hyperbola required. 

5. If straight lines be drawn from the point O, through 
the extremities CD of the conjugate axis, they will be the 
asymptotes of the hyperbola, whose property it is to ap- 
proach continually to the curve, and yet never to meet it. 

PROBLEM XII. 

In an hyperbola, any three of the four folloimng terms 
being given, viz. the transverse and conjugate diameters, 
an ordinate, and its abscissa, tofnd the fourth, 

CASE I. 

The transverse and conjugate diameters, and the two ab- 
scissas being known, to find the ordinate. 

RULE.t 

As the transverse diameter, 

Is to the conjugate ; 

So is the square root of the product of the two abscissas, 

To the ordinate required. 

* The sum of two lines drawn from the foci of an ellipse to any 
point in the curve, is equal to its transverse diameter. 

In like manner the difference of two lines drawn from the foci of 
any hyperbola to an}^ point in the curve, is equal to its transverse di- 
ameter, as is shown by the writers on conies. 

But the arcs intersecting each other in s, s. Sec. are described frorn 
the foci/ and F, and with the distances An, and Tin, whose differ- 
ence is AB, and therefore the points s, s, Sec. are in tlie curve of an 
by perhola. 

f Let ^ = transverse diameter, c = conjugate, a; = abscissa, and 



CONIC SECTIONS. 



109 



EXAJMPLES. 



1. In the hyperbola GAH, the transverse diameter is 
120, the conjugate 72, and the abscissa AF is 40 ; required 
the ordinate FH. 

A 




72 {conj.) : : 
6X x/ (160X40) 

= Yo "^ 

3x80 

—^r— =3X16=48 



V (160 X 40) : 
= -fv/ (160X40)=! 

-ordinate FH. 



s diameter is 24, the conjugate 21, and 
what is its ordinate? Ans. 14. 



120 (trans.) : 

72 X y/ (160X40) 
120 ' 

v/6400=|x80 = 

2. The trans versi 
the less abscissa 8 ; 

3. The transverse diameter of an hyperbola is 50, the 
conjugate 30, and the less abscissa 12 ; required the ordi- 
nate. ' Ans. 16.3658. 

CASE II. 

The transverse and conjugate diameters and an ordinate 
being given, to find the two abscissas. 

RULE.* 

As the conjugate diam.eter is to the transverse, 
So is the square root of the sum of the squares of the or- 
dinate and semi-conjugate. 

To the distance between the ordinate and the centre, or 
half the sum of the abscissas. 

y= ordinate. Then the general property of the curve Is t^ : c^ :: xx 
(t-{-x) : xf- ; and from this analogy aU the cases of this problem are 
deduced. 

Note — In the hyperbola, the less abscissa added to the axis, gi?es 
the greater. 

K 



110 CONIC SECTIONS. 

Then the sum of this distance and the semi-transverse 
will give the greater abscissa, and their difference the less 
abscissa required.* 



The transverse diameter is 120, the conjugate 72, and 
the ordinate 48 ; what are the two abscissas 1 

1296= square of the semi-conjugate* 
2^04:=^ square of the ordinate. 

S606(60= square root. 



00 

As 72 : 120 : : 60 
120 

72)7200(100=1 sum of the abscissas, 
72 60= semi-transverse. 

160= greater abscissa. 

AO:=less abscissa. 

2. The transverse and conjugate diameters are 24 and 
21 ; required the two abscissas to the ordinate 14. 

Ans. 32 and 8. 

3. The transverse being 60, and the conjugate 36 ; re- 
quired the two abscissas to the ordinate 24. 

Ans. 80 and 20. 



* This rule in species is L^ ^c''-\-y'±it= x, = greater 

or less abscissa, according as the upper or under sign is 
used. 



CONIC SEC5TI0N§. Ill 



CASE ni. 



The transverse diameter ^ the two abscissas and the ordirmte 
being given^ to find the conjugate, 

RULE. 

As the square root of the product of the two abscissas, 

Is to the ordinate ; 

So is the transverse diameter, 

To the conjugate.* 



1. The transverse diameter is 120, the ordinate is 48, 
and the two abscissas are 160 and 40 ; required the conju- 
gate. 

160 
40 



64 

00 
As 80 : 48 : : 120 ^Ae transverse a^is* 

48 

960 
480 



80)5760 



72 = conjugate required. 



* This rule, expressed algebraically, is ty-r- y/xx(txx) 
=:c= conjugate diameter* 



112 COXIC SECTIONS. 

2. The transverse diameter is 24, the ordinate 14, and 
the abscissas 8 and 32 ; required the conjugate. 

Ans. 21. 



CASE IV. 

The conjugate diameter, the ordinate and two abscissas 
being given, to Jind the transverse. 



RULE. 

1. Add the square of the ordinate to the square of the 
semi-conjugate, and find the square root of their sum. 

2. Take the sum or difference of the semi-conjugate and 
this root, according as the less or greater abscissa is used, 
and then say, 

As the square of the ordinate. 
Is to the product of the abscissa and conjugate ; 
So is the sum, or difference, above found, 
To the transverse required. 



1. The conjugate diameter is 72, the ordinate is 48, and 
tlie less abscissa 40 ; what is the transverse ? 



Here ^48^ +362 = V2304 + 1296=v/ 3600=60: 
And 60 + 36=96. 

Also 72 X ^0 =2880 =product of the abscissa and conju- 
gate. Whence, 



ex 



* This rule in algebraic terms is — X ( V ic^ +y3 =^c) 
=^= transverse diameter* 



CONIC SECTIONS* 113 



As 2304 : 2880 : : 96 
96 



17280 
25920 



2204:)276A80{l20z=ztrunsverse required, 
2304 



4608 
4608 



2. The conjugate diameter is 21, the less abscissa 8, and 
its ordinate 14; required the transverse. Ans. 24. 

3. Required the transverse diameter of the hyperbola, 
whose conjugate is 36, the less abscissa being 20, and or- 
dinate 24. Ans. 60. 



PROBLEM XIII. 

To find the length of any arc of an hyperbola, beginning 
at the vertex. 



RULE.* 

1. As the transverse is to the conjugate, so is the conju- 
gate to the parameter. 



'* Demon, Let i= semi -transverse axe, c— serai-conju- 

t^ 
gate, a:=ordinate, and 2/=abscissa. Then will 2/X(l + ^ 

11 «'* 4- Ti V 5 &c.)= length of the 

y 40c8 y ^ 112^^2 ^ ' ^ s 

^rc, as is shown by the writers on fluxions. 

t 2c , , 

Butx—~^c^-\-y^ — a, and — =parameter=p, by the 

K2 



114 COXIC SECTIONS. 

2. To 19 times the transverse, add 21 times the param- 
eter of the axis, and multiply the sum by the quotient of 
the abscissa divided by the transverse. 

3. To 9 times the transverse, add 21 times the parame- 
ter, and multiply the sum by the quotient of the abscissa 
divided by the transverse. 

4. To each of the products, thus found, add 15 times the 
parameter, and divide the former by the latter ; then this 
quotient being multiplied by the ordinate will give the 
length of the arc nearly. 

EXAMPLES. 

1. In the hyperbola GAE, the transverse diameter is 80, 
the conjugate 60, the ordinate GH 10, and the abscissa 
AH 2.1637 J required the length of the arc AG. 




nature of the curve. Consequently the rule becomes (15/)+ 
l9t-\-2lp X y)-^(15^+ 97+21^ X y) X y = {Ibpt + l^tx 

2x 



+ 21px)-^(irmt-i-9tx-\-21px)Xy=yX : 1 + ? 



2ar2 

- — -— ,&c. 
Sp 5/> 

which by substituting the values of a: and p^ and expanding 
the terms, gives a series, agreeing nearly in the first three 
terms with the former ; and therefore, the rule is an ap- 
proximation. 

If <=semi-transverse, c= semi-conjugate, and 7/=ordi- 
nate drawn from the end of the rcc|uircd arc ; then yX (1 + 
ty ^+ 4<f Srf t' + 4/c2+ 8c' 5/ 

6^A— ^^.--n + ^2+40^ .j^c,d:c.)==length 



of the arc. 



CONIC SECTIONS. 115 

60x60 3x60 
Here80:60 ; : 60 : qq = — ^ =3X15 = 45 = 

parameter, 

2.1637 

An<?(80xi9x45x21)X—g^=1520 + 945X. 02704 

=2465 X .02704=66.6536. 

, 2.1637 

^Zso (80x 9 + 45X 21)X —Q^=720 + 945X .02704= 

1665 X. 02704=45.0216. 



Whence 675 + 66.6536 -f- 675 + 45.0216 = 741.6536-f- 
720.0216=1.03004; and 1.03004 X 10= 10.3004=Zew^^A 
of the arc required. 

2. The transverse diameter of an hyperbola is 120, the 
conjugate 72, the ordinate 48, and the abscissa 40 : re- 
quired the length of the curve. Ans. 62.6496. 

3. Required the whole length of the curve of an hyper- 
bola, to the ordinate 10 ; the transverse and conjugate axes 
being 80 and 60. Ans. 20.6006. 

PROBLEM XIV. 

To find the area of an hyperbola, the transverse, conjugate ^ 
and abscissa being given, 

RULE.* 

1. To the product of the transverse and abscissa, add ^ 
of the square of the abscissa, and multiply the square root 
of the sum by 21. 



* Demon, Let ^= transverse diameter, c= conjugate, x= 
abscissa, ?/=ordinate, and z= ^^^^ . Then it is well known 

that 4:xy X (— _-i — ^ _-i z^ — —^ -s^ ^ >> _ 

^ 3 1.3.5. ^ 3.5.7. ^ 5.7.9. ^ ' ^^'^ " 



area of the hyperbola. 



116 COyiC SECTIONS. 

2. Add 4 times the square root of the product of the 
transverse and abscissa, to the product last found, and di- 
vide the sum by 75. 

3. Divide 4 times the product of the conjugate and ab- 
scissa by the transverse, and this last quotient multiplied 
by the former will give the area required nearly. 



EXAMPLES. 

In the hyperbola GAF, the transverse axis is 30, the 
conjugate 18, and the abscissa or height AH is 10; what 
is the area ? 



ty 
But — ==c=coniugate axis, by the nature of the 

hyperbola. Consequently the expression for the rule= - 



And this thrown into a series will very nearly agree with 
the former ; which shows the rule to be an approximation. 

Q. E. I. 

Rule 2. If 2y, 2y= bases, v, and r their distances from 
the centre, and the other letters as before, then will vy — 

tc ax + cv 
vy — —r X hyp. loff. of - — ; = area of the frustrum of the 

hyperbola. 

Rule 3. If t be put = transverse axis, c = conjugate, 
and X = abscissa, the area of a segment of an hyperbola, 

; ut off by a double ordinate will be = _ 4 -\- 



15 



X - very nearly. 



CONIC SECTIONS. 

A 



117 




ggrg 217(30X10 + 4X1 0^)= 21 y/ SOO + 500 ^ 7 = 
21 V300 + 71.42857=21 ^371.42857 =21 X 19.272= 
404.712. 



And (4V30X 10 + 404.712)^75 = (4V300+404.712) 
H-75= (4 X 17.3205 -{- 404.712)-^75 = (69.282 + 404.712) 
—75=473.994^-75=6.3199. 



Whence 



18X10X4 



X 6.3199 = 



18X4 



X 6.3199 = 6 



30 3 

X 4X 6.3199=24X 6.3199=151.6776=area required. 

2. The transverse diameter is 100, the conjugate 60, and 
the less abscissa 50 ; what is the area of the hvperbolal 

Ans. 3220.363472, 

3. Required the area of the hyperbola to the abscissa 
25, the two axes being 50 and 30. Ans. 805.0909, 



MENSURATION OF SOLmS. 



DEFINITIONS. 



1. The measure of any solid body, is the whole capacity 
or content of that body, when considered under the triple 
dimensions of length, breadth, and thickness. 

2. A cube whose side is one inch, one foot, or one yard, 
&c. is called the measuring unit ; and the content or solidi- 
ty of any fi{Tiire is computed by the number of those CUbes 
contained in that figure. 

3. A cube is a solid contained by six equal square sides. 




4. A parallelopipedon is a solid contained by six quad- 
rilateral planes, every opposite two of which are equal and 
parallel. 




5. A prism is a solid whose ends are two equal, parallel, 
and similar plane figures, and whose sides are parallelo- 
^ams. 



OP SOLIDS. 119 

Note. — ^W'hen the ends are triangles it is called a trian- 
gular pris?n ; when they are squares, a square prism ; when 
they are pentagons, a pentagonal prism, &;c. 



6. A cylinder is a solid described by the revolution of a 
right angled parallelogram about one of its sides, which 
remains fixed. 



7. A ^pyramid is a solid whose sides are all triangles 
meeting in a point at the vertex, and the base any plane 
figure whatever. 

Note. — When the base is a triangle, it is called a trian' 
gular pyramid; when a square, it is called a square or 
quadrangular pyramid ; when a pentagon, it is called a 
pentagonal pyramid, &c. 




8. A sphere is a solid described by the revolution of a 
semicircle about its diameter, which remams fixed. 




9. The centre of a sphere is a point within the figure, 
everywhere equally distant from the convex surface of it. 

10, The diameter of the sphere is a straight line passing 

* The definition of a cone has been given already. 



120 MENSUBATION 

through the centre, and terminated both ways by the con- 
vex superficies. 

11. A circular spindle is a solid generated by the revo- 
lution of a segment of a circle about its chord, which r'^ 
mains fixed. 



12. A spheroid is a solid generated by the revolution of 
a semi-ellipsis about one of its diameters, which is consid- 
ered as quiescent. 

The spheroid is called prolate, when the tevolution is 
made about the transverse diameter, and oblate when it is 
made about the conjugate diameter. 




13. Elliptic, parabolic, and hyperbolic spindles, are 
generated in the same manner as the circular spmdle, the 
double ordinate of the section being always fixed or quies- 
cent. 

14. Parabolic and hyperbolic conoids, are solids formed 
by the revolution of a semi-parabola or semi-hyperbola 
about its transverse axis, which is considered as quiescent. 



15. The segment of a pyramid, sphere, or of any other 
solid, is a part cut oflf from the top by a plane parallel to 
the base of that solid. 

16. A frustrum or trunk, is the part that remains at the 
bottom, after the segment is cut off. 

17. The zone of a sphere, is that part which is inter- 



OP SOLIDS. 121 

cepted between two parallel planes ; and when those planes 
are equally distant from the centre, it is called the middle 
zone of the sphere. 

18. The height of a solid is a perpendicular, drawn 
from its vertex to the base or plane on which it is supposed 
to stand. 

PROBLEM I. 

Tojind the soliditi/ of a cube, the height of one of its sides 
being given. 

RULE.* 

Multiply the side of the cube by itself, and that product 
again by the side, and it will give the solidity required. 



1. The side AB, or BC, of the cube ABCDFGHE, is 
25.5: what is the solidity? 




• Demon^ Conceive the base of the cube to be divided into a num- 
ber of little squares, each equal to the superficial measuring unit. 

Then will those squares be the bases of a like number of small 
cubes, which are each equal to the solid measuring unit. 

But the niunber of little squares contained in tiie base of the cube 
are equal to the square of the side of that base, as has been shown al- 
ready. 

And consequently, the number of small cubes contained in tlie 
whole figure, must be equal to the square of the side of tlie base mul- 
tiplied by the height of that figure ; or, which is the same thing, 



122 MENSUBATION 

Here AB^ =^22l5| » =25.5 X 25.5 X 25.5= 25.5 X 650.25 
= 16581.375 the answer, or content of the cube. 

2. The side ol a cube is 15 inches : what is the solidity ? 

Ans. \ft. Win, 5pa. 

3. What is the solidity of a cube whose side is 17.5 
inches? Ans. S.1015 feet. 

PROBLEM II. 
To find the solidity of a parallelopipedon. 

RULE.* 

Multiply the length by the breadth, and that product 
again by the depth or altitude, and it will give the solidity 
required. 



1. Required the solidity of a parallelopipedon ABCD 
FEHG, whose length AB is 8 feet, its breadth FD 4^ feet, 
and the depth or altitude AD 6| feet ? 




the square of the side of the base multiplied by the base, is equal to 
the solidity. Q. E. D. 

Note. — The surface of the cube is equal to six times the squzire of 
its side.- 

* The reason of this rule, as well as of the following ones for the 
prism and cylinder, is the same as that for tlie cube. 

iVi9^e. — The surface of the parallelopipedon is equal to the sum of 
the areas of each of its sides or eads. 



OF SOLIDS. 123 

Here ABx ADx FD = 8 x 6.75 X 4.5 = 54x 4.5 = 243 

solid feet, the contents of the parallel opipedon required, 

2. The length of a parallelopipedon is 15 feet, and each 
side of its square base 21 inches : what is the solidity? 

Ans 4.5.9375 feet. 

3. What is the solidity of a block of marble, whose 
length is 10 feet, its breadth 5f feet, and the depth 3^ 
feet] Ans. 201.25/ee^ 

PROBLEM ni. 
To find the solidity of a prism* 

RULE.* 

Multiply the area of the base into the perpendicular 
height of the prism, and the product will be the solidity. 

EXAMPLES. 

1. What is the solidity of the triangular prism ABCF 
ED, whose length AB is 10 feet, and either of the equal 
sides, BC, CD, or DB, of one of its equilateral ends BCD, 
2ifeetl 

E 




* The surface of a prism is equal to the sum of the areas of the two 
ends and each of its sides* 



124 MENSURATION 

* Hereix2.5^X v/3=^x6.25x V3=1.562oX v/3= 
1.6825 X 1.732 = 2.70625=area of the base BCD. 
2.5 + 2.5 + 2.5 7.5 

Qj.^ o ""2" ~ ^*^^ ~ ^ ^^''^ ^ «iae5, 

BC, CD, DB, o/f^e triangle CDB. 

A/w^ 3.75—2.5=1.25, .-. 1.25, 1.25 and 1.25z=3 dijer- 
ences. 

Mlience s/ 3.75 X 1.25 X 1.25 X 1.25= V 3.75 X 1.252 = 
V'7.32421875=2.7063=area of the base as before, 

And 2.7063 X 10=27.063 solid feet, the content of the 
prism required. 

2. What is the solidity of a triangular pripm, whose 
length is 18 feet, and one side of the equilateral end 1^ 
feet ? Ans. 17.5370265 feet. 

3. Required the solidity of a prism whose base is a hexa- 
gon, supposing each of the equal sides to be 1 foot 4 inches, 
and the length of the prism 15 feet. Ans. 69.282ft. 

PROBLEM IV. 

To find the convex surface of a cylinder. 

RULE.t 

Multiply the periphery or circumference of the base, by 
the height of the cylinder, and the product will be the con- 
vex surface required. 

EXAMPLES. 

1. What is the convex surface of the right cylinder 
ABCD, whose length BC is 20 ieet, and the diameter of its 
base AB 2 feet ? 

* See Notes to Proh. III. Cor. 2. p. 56. 

t Demon. If the periphery of the base be conceived to move in a 
direction parallel to itself, it will grcnerate the convex superficies of 
the cylinder ; and, therefore, tlic said pcrijihery being- multiplied by 
the length of the cylinder, will be equal to that superficies. Q. E. D. 

Note. If twice the area of either of tlie ends be added to the con- 
rex surface, it will give the whole surface of the cylinder. 



OF SOIilBS. 



123 




Here S.l4:Wx2 = 6.28S2=periphery of the base AB. 
And 6.2832X20=125.6640 square feet, the convexity 
required. 

2. What is the convex surface of a right cylinder, the 
diameter of whose base is 30 inches, and the length 60 
inches? Ans. 5654.88 inches* 

3. Required the convex superficies of a right cylinder, 
whose circumference is 8 feet 4 inches, and its length 14 
feet- Ans. 116.666, cf-c. /eet 

PROBLEM V. 

To find the solidity of a cylinder. 

RULE.* 

Multiply the area of the base by the perpendicular 
height of the cylinder, and the product will be the solidity. 

* The four following eases contain all the rules for finding the su- 
perficies and solidities of cylindric ungnlas. 

1. When tJi-e section is parallel to the axis of the cylinder. 

F 




L2 



126 



MENSURATION. 



EXAMPLES. 



1. What is the solidity of the cylinder ABCD, the di- 
ameter of whose base AB is 30 inches, and the height BC 
50 inches ? 



Rule 1. Multiply the length of the arc line of the base by the height 
of the cylinder, and the product wiU be the curve surface. 

2. Multiply the area of the base by tlie height of the cylinder, and 
the product will be the solidity. 

II. When the section passes ohliquely through the opposite sides of 
the cylinder. 




Rule 1. Multiply tlie circumference of the base of the cylinder by 
half the sum of the greatest and least lengths of the ungulei, and the 
product wiU be the curve surface. 

2. Multiply the area of tlie base of the cylinder by half the sum of 
the greatest and least lengths of the iingula, and the product will be 
the solidity. 

III. When the section passes tkrovgh the hase of the cylinder, and 
one of its sides. 




Rule 1. Multiply the sine of halfthe arc of the base by the diameter 
of the cylinder, and from tliis product subtract tlie product of the arc 
and cosine. 

^. Multiply ihe difference thus found, by tlie quotient of the height 
divided by the versed sine, and tlie product will be the curve surface. 



OF SOXIDS. 



12T 




* Here .7854 X 302=. 7854 X 900=706.86=:area of the 
base AB. 



3. From f of the cube of the right sine of half the arc of the base, 
subtract the product of the area of the base and the cosine of the said 
half arc. 

4. Multiply the difference, thus found, by the quotient arising- from the 
height divided by the versed sine, and the product will be the solidity. 

IV. When the section passes obliquely through loth ends of the cyU 
inder. 

c 




Rule 1. Conceive the section to be continued, till it meets the side 
of the cylinder produced ; then say, as the difference of the versed 
sines of half the arcs of the two ends of the ungula, is to the versed 
sine of half the arc of the less end, so is the height of the cylinder to 
the part of the side produced. 

2. Find the surface of each of the ungulas, thus formed, by Case III. 
and their difference will be the surface required. 

3. In like manner find the solidities of each of the ungideis, and 
their difference will be the solidity required. 

* In working the examples in this and the following rules, .7854 
is used for the area, and 3.143 6, the circumference of a circle whose 
diameter is 1 ; where greater accuracy is required, .7853981634 may 
be used for the area, and 3.14159265359 for the circumference. See 
Note to Prob. IX. Superficies. 



126 



MENSURATION 



35343 



And! 06.86x50 = 353^3 cubic inches; o r ^^^^ =20.4531 
solid feet, the answer required, 

2. What is the solidity of a cylinder whose height is 5 
feet, and the diameter of the end 2 feet ? 

Ans. 15.708 feet, 

3. What is the solidity of a cylinder whose height is 20 
feet, and the circumference of its base 20 feet also ? 

Ans. 636.64 feet. 

4. The circumference of the base of an oblique cylinder 
is 20 feet, and the perpendicular height 19.318 ; what is the 
solidity? '^ Ans. 614.93 /ee^ 

PROBLEM \1. 

To find the convex surface of a right cone. 
RULE.* 

Multiply the circumference of the base by the slant 
height, or the length of the side of the cone, and half the 
product will be the surface required. 



* Demon. Let AB=a, Bc=^h, 3.1416=p, and ED=y. 

hy 
Then a : b :: y : — =DC; andpy=circumferenceof the 

circle ED. 

But^X — =fluxion of the surface of CED, and its 
a 
phy^ pha 

fluent=-2^ which, when y=:a, becomes-g- = convex sur- 
face of the whole cone. Q. E. D. 

I To find the surface of a right pyramid. 

Rule. Multiply the perimeter of the base by the lenjlh of the side, 
or slant height of the cone, and half tlie product will be the surface 
required. 



129 



EXA3IPLES. 

1. The diameter of the base AB is 3 feet, and the slant 
height AC or BC 15 feet ; required the convex surface of 
the cone ACB. 

C 



Here 3.1416 X ^=QA2^Sz=circumference of the base AU» 

(22.5 + 15.75) X 26 

Here- ^ =22.5 + 15.75 X 13=38.25x 

lS^=4i91 ^25= convex surface required* 



Then T:p:: h(BC) : CD ; and, by division, P— p :p:: 
pk ph 

b — CD{h) : CD==-— - ; but p X (^ + -^—- = twice the con- 
vex surface of the whole cone, by the last rule ; and also p X 
ph 

= twice the convex surface of the part ECD. There- 

p — p ^ 

forepx (h+ -l^)—px I.—=hT-\-T^X-l— = Ap + 

p — p p — p p — p 

hp=F-{-pXh= twice the convex surface of the frustrum 

ABDE ; and the half thereof is i^-^P)^^ which is the same 

2 
as the rule. Q. E. D. 

Tojind the surface of the frustrum of a right pyramid. 

Rule. Multiply the sinn of the perimeters of the ends by the slant 
height, and half the product will be the surface required. 



180 MENSURATION 

EXAMPLES. 

1. In the frustrum ABDE, the circumferences of the two 
ends AB and DE are 22.5 and 15.75 respectively, and the 
slant height BD is 26 j what is the convex surface ? 

C 



^^^ 9.4248X 15 ^ 141.3720 ^^Q^^Qg .^are /eef, the 
2 Z 

convex surface required, 

2. The diameter of a right cone is 4.5 feet, and the 
slant height 20 feet ; required the convex surface. 

Ans. 141.372 /€e^ 

3. The circumference of the base is 10.75, and the slant 
height 18.25 ; what is the convex surface? 

Ans. 98.09375. 

PROBLEM VII. 

To find the convex surface of the frustrum of a right 
cone. 

RULE.* 

Multiply the sum of the perimeters of the two ends, by 
the slant height of the frustrum, and half the product will 
be the surface required. 

* Demon. Let the perimeter of the circle AB= P, that of DE=;), 
BD=A, and tlie rest as in the last problem. 



OP SOLIDS. 131 

1. What is the convex surface of the frustrum of a right 
cone, the circumference of the greater end being 30 feet, 
that of the less end 10 feet, and the length of the slant side 
20 feet ? Ans. 400 feet. 

2. What is the convex surface of the fmstrum of a right 
cone, the diameters of the ends being 8 and 4 feet, and the 
length of the slant side 20 feet? Ans. 376.992 feet. 

3. If a segment of 6 feet slant height be cut off a cone 
whose slant height is 30 feet, and circumference of its base 
10 feetj what is the surface of the frustrum? 

Ans. 144 feet. 



PROBLEM Vm. 
7b find the solidity of a cone or pyramid. 

RULE.* 

Multiply the area of the base by one third of the perpen- 
dicular height of the cone or pyramid, and the product will 
be the solidity. 

* Demon, Let sc=:a, cs=x, and A=area of the base 
of the cone acb. 

Then a^(cs^) : x^(cs^) : : ab^ : ed^ (by sim. As) : : a : 

— , (=area of the circle ed) because all the circles are 
as the squares of their diameters. 

AX^ ' 

But — Xa:= fluxion of the cone ecd, and its fluent= 

t^ , which, when x=a, becomes f:^ = aX — for the solid- 
3a2 3 3 

ity of the whole cone. Q. E. D. 

In the pyramid cede it will be a^(cs-) : x^(cs^) :: ce^ 

: ce^ :: ed^ : eo^ (by sim. as) ; : a (area of the base eb) : 

— (area of the polygon eh) because all similar figures are 



as the squares of their like sides. 



133 



MENSURATION 



EXAMPLES. 



1. Required the solidity of the cone ACB, whose diam- 
eter AB is 20, and its perpendicular height CS 24. 
C 




Here .7854 X 202 = .7854x 400=314.16 = area of the 
base AB. 

24 
And 314.16 X y = 314.16X 8 = 2513.28 = soZirfi^^/ re- 
quired. 

2. Required the solidity of the hexagonal pyramid ECBD, 
each of the equal sides of its base being 40, and the per- 
pendicular height CS 60. 

C 




But — Xa:= fluxion of the pyramid cco&, and its cor- 

rect fluent = aX — the same as in the cone : and this rule 

3 
is general, let the figure of the base be what it will. 



OF SOLIDS. 133 

Here 2.598076 {multiplier when the side is 1) X 40^= 
2.598076 X 1600=4156.9216=area of the base. 
60 
And 4156.9216 X t.- =4156.9216 X 20 =83138.432 so- 

o 

lidity required, 

3. Required the solidity of a triangular pyramid, whose 
height is 30, and each side of the base 3. Ans. 38.97117. 

4. Required the solidity of a square pyramid, each side 
of whose base is 30, and the perpendicular height 20, 

Ans. 6000. 

5. What is the solidity of a cone, the diameter of whose 
base is 18 inches, and its altitude 15 feet? 

Ans. 8,83575 /eel. 

6. K the circumference of the base of a cone be 40 feet, 
and the height 50 feet ; what is the solidity ? 

Ans- 2122.1333/ecf. 

7. What is the content of a pentagonal pyramid, its 
height being 12 feet, and each side of its base 2 feet? 

Ans. 27.5276. 

PROBLEM IX, 

Tojind the solidity of afrustrum of a cone or pyramid* 
RULE,* 

1. For the fruslrum of a cone, the diaineters, or circum- 
ferences of the two ends, and the height being given^ 

* Demoiu First let d= diameter ab, t^=ED, /7=.7854, 
^=ss= the height of the frustrum abde of the cone. See 
the lastfgures. 

Then d : d :: cs : cs, and d — d : d z: cs — cs (h) : 

^^ , > , . , ^ po^ ^h ^ 

—;^=cs= height of tne cone edc. But — x (^+ - — -,) 

= solidity of the whole cone acb, and-^X -5= the 

o D — a 

aolidity of the cone ecd. Therefore -^X(k+ zr~^ } 
M 



134 MENSURATION 

Add together the square of the diameter of the greater 
end, the square of the diameter of the less end, and the 
product of the two diameters ; multiply the sum by .7854, 
and the product by the height ; ^ of the last product will 
be the solidity. Or, 

Add together the square of the circumference of the 
greater end, the square of the circumference of the less 
end, and the product of the two circumferences ; multiply 
he sum by .07958, and the product by the height ; | of the 
last product will be the solidity. 

II. For the frustrum of a pyramid whose sides are regu- 
lar polygons. 

Add together the square of a side of the- greater end, 
the square of a side of the less end, and the product of 



,»«? dh . , ^ jyh „ dh . p 

Dh . p D^ — d^ hp , „ ,2 7N hp 

the solidity of the frustrum ABDE, \\-hich is the same as the rule. 
• And, since the circumferences of circles have the same ratio that 
their diameters have, if C be put for tlie circmnference of the greater 
end, c=that of the less end, and ^=.07958, the demonstration of the 
rule, when the circumferences are given, will differ in nothing from 
the above. 

Again, for the polygon, let S=ED, s=ed, and wi= proper multi- 
plier in the table of polygons ; then S ; s : : CS : Cs, and S— « is:: 

hs 
C9—C8 (h) : • 

^ ' S — 8 

But ms^ and ms^ are the areas of polygons whose sides 

ms^ hs 

are s and s respectively. And therefore -^-|-(/i4- ) 

ms^ hs 5 h 

3-X;iI^=(wW^+^s2— 7/i52 X-^3^X -g =(OTS2-f 771^ 

h mh 

+ mss) X -3-— (s^ +5^ +ss)x -g-= solidity of the frustrum 

CEDBb which is the same as the rule. 



OF SOtlBS. 1^5 

these two sides ; multiply the sum by the proper number 
in the table, Prob. YIII. of Superficies, and the product by 
the height : ^ of the last product will be the solidity. 

Note, — When the ends of the pyramids are not regular 
polygons. Add together the areas of the two ends and 
the square root of their product ; multiply the sum by the 
height, and \ of the product will be the solidity. 

EXAMPLES, 



1. What is the solidity of the frustrum of the cone 
EABD, the diameter of whose greater end AB is 5 feet, 
that of the less end ED, 3 feet, and the perpendicular 
height S5, 9 feet] 




(52^.82^5X3)X.7854X9 
3 " 



346.3614 



3 



==115.4538 solid 



feet^ the content of the frustrum, 

2. What is the solidity of the frustrum eEDB& of an 
hexagonal pyramid, the side ED of whose greater end is 
4 feet, that eh of the less end 3 feet, and the height S^, 
9 feet] 




136 MENSURATION 

(4«4. 32-f 43^3) X 2.598076 X 9 865.159305 

^ ^ = 3 =288.386436 

golidjeet, tJie solidity required* 

3. What is the solidity of the frustrum of a cone, the 
diameter of the greater end being 4 feet, that of the less 
end 2 feet, and the altitude 9 feet? Ans. 65.9736. 

llie foUowing- cases contain all the rules for finding the superficies 
and solidities of conical ungulas. 

1. When the section passes through the opposite extremities of the 
ends of the frustrum. 



Let D=AB the diameter of the greater end ; c?=cd, the 
diameter of the less end ; A=perpendicular height of the 
frustrum, and n=.7854. 
d^ — d-Jjid ni)h 

Then — j — X -3-= solidity of the greater elliptic 

ungula ADB. 

ly^Dd — d^ ndh 

^ — X -g-=solidity of the less ungula acd. 



(p^— <^)2 nk 



— 7 — ^X -0-= difference of these hoofs. 

n D-{-d 

And^-^v'4A2+(D — rf2) X (d3_-^v'd^) =curve 

surface of adb. 

II. When the section cuts off part of the base, and makes the angle 
DrB less than the angle CAB. 




OF SOLIDS, 137 

4. What IS the solidity of the fnistrum of a cone, the 
circumference of the greater end being 40, that of the less 
end 20, and the length or height 50 1 Axis, 3713.7333. 



Let s= tabular segment, whose versed sine is sr—D, s — 
tab. seg. whose versed sine is Br — (n — d)-~d, and the 
other letters as above^ 

B^ B*" 4A 

Then (sx D^-— 5X d^X ==^,V =,)X -^ = 

Br — D — a Br — d — a d — a 
solidity of the elliptic hoof efbd. 

1 d' 

"^"^ ^Hd "^ ^^^ "^ ^^ — ^^^ ^ (^^^* ^^^ — "d^ ^ 
^ X (d + d) — xr Br 

d AT ^ '^ d^ZIZr^ ^^^' ^^ ^^^ circle ab, whose 

d — Ar 
height is dX — j — )= equal convex surface of efbd. 

III. Wh£n the section is parallel to one ef the sides of the frustruni. 



Let A =; area ef the base FBE, and the other letters as before, 

A X D , — 

Then (- — ^ — f£?V(B^ — d) x d) X P=solidityof the para- 
bolic hoof EFBD. 

1 

V^^X D—<i)= convex surface of efbd. 

IV. When the section cuts off part of the base, and maJces the angle 
DrB greater than the angle CAB. 

M2 




138 MENSURATION 

5. What is the solidity of the fnistrum of a square pyra- 
mid, one side of the greater end being 18 inches, that of 
the less end 15 inches, and the height 60 inches ? 

Ans. 16380 inches. 

6. What is the solidity of the frustrum of an hexagonal 
pyramid, the side of whose greater end is 3 feet, that of 
the less end 2 feet, and the length 12 feet? 

Ans. 197.453776/ert. 

PROBLEM X. 
To find the solidity of a cuneus or wedge. 

RULE.* 

Add twice the length of the base to the length of the 
, and reserve the number. 



Let the area of the hyperbolic section EDF= A, and the area of the 
circular seg. EBF=fif. 

rr<u i^ dXnr 

h^ (aX D— AX — \ — )= solidity of the hyper- 

bolic ungula efbd. 

1 <P 

And 2^ y/4.¥-{-{Tf — rf)^x(cir. seg. ebf— -r X 

^r—i (j>—d) Br _r r 

V == = curve surface of efbd. 

Br^— D — d Br — d — d 

Note. — ^The transverse diameter of the hpy. seg. = 

dxcr Br 

p— rf_Br ^"^ ^^^ co"J"gate=<f v^ p__^_^^^ , from which its 
area may be found by the former rules. 

* Demon. When tlie leng^th of the base is equal to half of the 
wedge, the wedge is evidently equal to half a prism of the eame 
l»«e and altitude. 



OF SOLIDS. 139 

Multiply the height of the wedge by the breadth of the 
base, and this product by the reserved number; } of the 
last product will be the solidity. 



EXAMPLES. 

1. How many solid feet are there in a wedge, whose 
base is 5 feet 4 inches long, and 9 inches broad, the length 
of the edge being 3 feet 6 inches, and the perpendicular 
height 2 feet 4 inches 1 



(64X2 + 42) X28X9 _ (128-|-42) X28X9 _ 
— g — 

170X14X3 = 7140 solid 



Here 

170X28X9 170X28X3 



6 ~ 2 

inches. 

And 7140-^1728 = 4.1319 solid feet, the content re- 
quired. 

And according as the edge is shorter or longer than the base, the 
wedge is greater or less than half a prism, by a pyramid of tlie same 
height and breadth at the base with the wedge, and the length of 
whose base is equal to the difference of the lengths of the edge and 
base of the wedge. 

Therefore, let the length of the base BC=L ; the length of the 
edge EF=Z; the breadth of the base BA=h; and the height of the 
wedge EP=A ; and we shall have by the former rules 

2 - 3 2 3 6 



140 MENSURATION 

2. The length and breadth of the base of a wedge are 
35 and 15 inches, and the length of the edge is 55 inches : 
what is the solidity, supposing the perpendicular height to 
be 17.14508 inches? Ans. 3.1006/eef. 



PROBLEM XI. 

To find the solidity of a prismoid, 

RULE.* 

To the sum of the areas of the two ends add four times 
the area of a section parallel to and equally distant from 
both ends, and this last sum multiplied by ^ of the height 
will give the solidity. 

Note. — The length of tiie middle rectangle is equal to 
half the sum of the lengths of the rectangles of the two 
ends, and its breadth equal to half the sum of the breadths 
of those rectangles. 



* Demon. The rectangular prismoid is evidently comjxjsed of 
two wedges, whose heights are equal to the height of the pris. 
moid, and their bases its two ends. Wherefore, by the last 

k 

problem its solidity will be=(2L + Zx b + 2Z+lX 6)X g, 

L-fZ, B + Z>, 

which, by putting m = —^ and 7/i = — ^ — becomes 

h , . 

BL + Z>Z + 43rmX ^; which is the rule, as was to be shown. 

The solidities of the two parts, commonly called the imgulas, or 
hoofs, into which the frustrum of a rectangular pyramid is divided, 
may be found by the last two rules, as they are only composed of 
wedges and prismoids, 

A very elegant demonstration of this rule for the prismoid may be 
seen in Simpson's Fluxions, p. 179, 2</ Edition. 

If the bases of the prismoid are dissimilar rectangles, of which 
L, Z and M, m are corresponding dimensions, and h the height; 

Then (2 -}- lZ.m -f 2Z + l.w) X i^= solidity. 



OF SdUDS. 141 



EXAMPLES. 



1. What is the solidity of a rectangular prismoid, the 
length and breadth of one end being 14 and 12 inches, and 
the corresponding sides of the other 6 and 4 inches, and 
the perpendicular 30^ feet. 



JTcre 14X 12-f 6 X 4=168 + 24=192=«ttffi of Me orea« 
ojT the two ends. 

14 + 6 20 
Also — 2 — "^ "o" =10=?e«^A of the middle rectangle* 

12-f4 16 
And g— =^=8=&rea<fM of the middle rectangle. 

Whence lOX 8X 4=80X 4=320=4 times the area of 
the middle rectangle. 

366 
Or (320 + 192) X -^ =512 X 61 = 31232 solid inches. 

And 31232^1728=18.074 solid feet, the content. 

2. What is the solid content of a prismoid, whose greater 
end measures 12 inches by 8, the less end 8 inches by 6, 
and the length, or height, 60 inches ? Ans. 2A5Sfeet. 

3. What is the capacity of a coal wagon, whose inside 
dimensions are as follow : at the top, the length is 81^, and 
breadth 55 inches ; at the bottom the length is 41, and the 
breadth 29^ inches ; and the perpendicular depth is 47^ 
inches? Ans. 126340.59375 cubic inches^ 



148 ME?7SURATION 

PROBLEM XIL 

To find the convex surface of a sphere* 

RULE.* 

Multiply the diameter of the sphere by its circumference, 
and the product will be the convex superficies required. 

Note. — The curve surface of any zone or segment will 
also be found by multiplying its height by the whole cir- 
cumference of the sphere. 



* Demon. Put the diameter BG=d, BA=j?, AC=y, BC=z ; and 
3.14J6=p. 

Then, since the triangles AOC and CED are similar, we shall 
d . . dx 

have CA (y) :co-^ :: ce (j) : cd(z)=~. But 2pyz is the gene- 
ral expression for the fluxion of any surface; and therefore, by 

dx ^ 
substituting k~ for its equal z, the fluxion will become pdx ; and 

consequently ^(Zjr= surface of any segment of a sphere whose height 
is X, and pdd=theLt of the whole spliere. Q. E. D. 

Cor. 1. The surface of a sphere is also equal to the curve surface 
of its circumscribing cylinder. 

Cor. 2. The surface of a sphere is also equal to four times the 
area of a great circle of it. 

1. To find the lunar surface included between two great circles of 
tlie sphere. 

Rule. Multiply the diameter into the breadth of the surface in the 
middle, and the product will be tlie superficies required. Or, 
As one right angle is to a great circle of the sphere ; 
So is the angle made by the two great circles, 
To the surface included by them. 

2. To find the area of a spherical triangle, or the surface included 
by the intersecting arcs of three great circles of tlie sphere. 

Rule. As two right angles, or 180'^, 

Is to a great circle of the sphere ; 

So is the excess of the three angles above two right angles. 

To the area of the triangle. 



OF SOLIDS. 
EXA3IPLES. 



143 



1. What is the convex superficies of a globe BCG, 
whose diameter BG is 17 inches? 




JE/ere 3.1416 X 17 X 17=53.4072x 17= 907.9224 sg^rc 
inches. 

And 907.9224H- 144=6.305 square feet, the anstcer. 

2. What is the convex superficies of a sphere whose di- 
ameter is 1^ feet, and the circumference 4.1888 feet? 

Ans. 5.58506 /ee^. 

3. If the diameter, or axis of the earth be 7957| miles, 
what is the whole surface, supposing it to be a perfect 
sphere? Ans. 198944286.35235 sq. miles. 

4. The diameter of a sphere is 21 inches ; what is the 
convex superficies of that segment of it whose height is 4^ 
inches? Ans. 296.8812 inches. 

5. What is the convex surface of a spherical zone, whose 
breadth is 4 inches, and the diameter of the sphere, from 
which it was cut, 25 inches? Ans. 314.16 inches. 

PROBLEM Xm. 

To find the solidity of a sphere or globe, 

RULE.* 

Multiply the cube of the diameter by .5236, and the 
product will be the solidity. 

* Demon. Put AD==a;, Dc=y, the diameter AB=d, and 
p=3.146. 



144 



KENSTJRATION 



EXAMPLES. 

1. What is the solidity of the sphere AEBC, whose di- 
ameter AB is 17 inches 1 




Here IT'' X 5236 = 17X 17X 17X .5236 = 289X17X 
5236=4913 X .5236 = 2572.4468 solid inches. 

And 2572.4468-^1728=1.48868 solid feet, the anstcer. 

2. What is the solidity of a sphere whose diameter is 1^ 
feet? Ans, 1,2 All feet, 

3. What is the solidity of the earth, supposing it to be 
perfectly spherical, and its diameter 7957^ miles? 

Ans. 263858149120 miles. 



Then, by the property of the circle, dx — x'^=y'. But 

the general expression for the fluxion of any solid is py'^x ; 

and therefore by writing dx — x^ for its equal y^, we shall 

have px X dx — a:^=pdxx — px'^x. The fluent of which is 

pdx^ pj? Spdx^ — 2px^x 

—X ^ = ^ =content of the segment cae. 

Spd^—2pd' 
And if d be substituted for x, it will become a 

pd^ 



6 



=<r X .5236, or .5236<f j which is the same as the rule. 

Coroll. A sphere, or globe, is equaJ to two-thirds of its circum- 
•cribing cylinder. 



OF soLros. 145 

PROBLEM XIV. 

To Jind the solidity of the segment of a sphere, 
RULE,* 

To three times the square of the radius of its base add 
the square of its height; and this sum multiplied by the 
height, and the product again by .5236, will give the so- 
lidity. Or, 

From three times the diameter of the sphere subtract 
twice the height of the segment, multiply by the square of 
the height, and that product by .5236 j the last product 
will be the solidity. 

EXA3IPLES. 

1. The radius Cn of the base of the segment CAD is 7 
inches, and the height An 4 inches ; what is the solidity ? 




* Demon. Let r= radius of the base of the segment, h 
=height of the segment, and the other letters as before. 

Then will {Uh^—21}?)x -^= solidity of the segment, as 
6 
is shown in the last problem. 

But since — r— =<:?, by the property of the circle we shall 

N 



146 MENSURAtlOl* 

Here (7=^X 3 + 42)x 4x .5236=(49x 3 + 42)x 4x .5236 
=(147 + 42)X4X.5236=(147 + 16)X4X. 5236 = 163X4 
X .5236=652 X .5236=341.3872 solid inches, the answer. 

2. What is the solidity of the seg-ment of a sphere, the 
diameter of whose base is 20, and its height 9 ? 

Ans. 1795.4244. 

3. What is the content of a spherical segment, whose 
height is 4 inches, and the radius of its base 8 ? 

Ans. 435.6352. 

4. What is the solidity of a spherical segment, the di- 
ameter of its base being 17.23368, and its height 4.5 ? 

Ans. 572.5566. 

5. The diameter of a sphere being six inches, required 
the solidity of the segment whose altitude is two inches. 

Ans. 29.3216 cubic inches. 

6. Required the solidity of a spherical segment, the 
height of which is 15, the diameter of the sphere being 18. 

Ans. 2827.44. 

PROBLE]\I XV. 

To find the solidity of a frustrnm or zone of a sphere^ 

RULE.* 

To the sum of the squares of the radii of the two ends, 
add one third of the square of their distance, or of the 



have ( ^ — 2k^)xf = Sr^ + A^ ^ ^— = solidity 

of the segment, which is the same as tlie rule. 

Or, if <7=diameter of the sphere, and A = hcight of the 
segment; then will .52367i-X (3^^— 2A)=solid:ty. 

* Demon. The difference between two segments of a sphere whose 
heights are H and h, and the radii of whose bases are R and r, will 

P 
by the last problem =g X (SR^ii-f h^ — 3i^h — }i^)=zone whose height 

ia n — h. And therefore by putting a for the altitude of the frustrum, 



OF SOLIDS. 147 

breadth of the zone, and this sum multiplied by the said 
breadth, and the product again by 1.5708, will give the 
solidity. 



1. What is the solid content of the zone ABCD, whose 
greater diameter AB is 20 inches, the less diameter CD 15 
inches, and the distance mn of the two ends 10 inchest 




103 
Here 101 + 7.5^ +-3-) X 10 x 1.5708= (100 + 56.254- 

33.33) XlO X 1.5708 = 189.58 X 10 X 1.5708= 1895,8 X 
1.5708=2977.92264 solid inches, the answer, 

2. What is the solid content of a zone, whose greater 
diameter is 24 inches, the less diameter 20 inches, and the 
distance of the ends 4 inches? Ans. 1566.6112 inches. 

3. Required the solidity of the n?iddle zone of a sphere, 
whose top and bottom diameters are each 3 feet, and the 
breadth of the zone 4 feet] Ans. 61.7848 feet. 



and exterminating H and h by the means of the two equations ^ 

= — T — andE— ^=a,we shall have (r2 4. r^-f-o") X "9"i which is the 
rule. 

If it be the middle zone of the sphere, the solidity will be = fZ' 4- §7<2) 
X .7854A ; where <^=diameter of each end, and A = it3 height. 



148 jiENsuRATioir 

PROBLEM XYL 
To find the solidityt of a spheroid* 

RULE.* 

Multiply the square of the revolving- axe by the fixed 
axe, and this product again by »5236, and it will give the 
solidity required* 

Where note that .5236 is=i of 3.1416. 

EXAMPLES. 

1. In the prolate spheroid ABCD, the transverse, or 
fixed axe AC is 90, and the conjugate, or revolving axQ 
DB is 70 : what is the solidity t 



* Denton. Let A,c=cty i>b=6, Ar=x, m=±ify and p= 
3.14159, &c. 

Then aF:¥:: xX (a — x) : -jX («-^ — x^)=ifhy the prop. 

erty of the ellipsis. 

• ' . ^ja 

xind therefore the fluxion of the solid {—pjfx)—-^X 

or 

(axx — x^x) ; and its fluent x t_ x (iaa^ — ixS) = segment 

pb'^ pab^ 

nxm. "Which, when ar=a becomes — j X (^o'— |a')= -«— 

= content of the whole spheroid. Q. E. D. 

If y be put=fixed axe, ?'== revolving axe, q==(J^u2 r')-r 
/2, andp=3.1416, &c. 

Then will prfy/l-{-iq=suTface of the oblate spheroid, 
9LnAprf^/l — i^=that of the prolate spheroid. 




149 



Here DB^X ACx .5236 = 70-' X 90 X .5236= 4900 X 90 X 
.5236=441000 X .5236 = 230907,6=soZif^iif?/ required, 

2. What is the solidit)^ of a prolate spheroid, whose fixed 
axe is 100, and its revolving axe 60? Ans. 188496. 

3. Vvhat is the solidity of an oblate spheroid, whose 
fixed axe is 60, and its revolving axe is 100? 

Ans. 314160. 

PROBLEM XVn. 

To find the content of the middle frustnim of a spheroid, 
its length, the middle diameter, and that of either of the 
ends, being given. 

CASE I. 

When the ends are circular, or parallel to the revolving 



RULE.* 

Te twice the square of the middle diameter add the 
square of the diameter of either of the enda, and this sum 



* Demon. Let Ao=:cr, jyo^=h, sn^=h, no=c^ ro=x, re~ 
jf, and jp= 3.14159, &c. 

Then a2 : h^ iia^—x^ i^x(a^-^s;^)—b^--——^ 



y^ by tlie property of the ellipsis. 

b^x(a^^ 

And also a^ :b^ :: a^ — c^ i ^ 

l^t^^a^^hK) 

N2 



n 



zh^; or o*= 



150 MENSURATION^ 

multiplied by the length of the frustrum, and the product 
again by .2618, will give the solidity. 
Where note that .2618=yV of 3.1416. 



1. In tlie middle frustrum of a spheroid EFGH, the 
middle diameter DB is 50 inches, and that of either of the 
ends EF or GH is 40 inches, and its length nm 18 inches : 
what is its sohdity] 

D 



A"r-»r 




Heie (oQ^X 2 + 402)x 18x .2618 = (2500X 2 + 16Q0)x 
18 X. 2618 = (5000 + 1600) X 18 X. 2618= 6600 X 18 X 
.2618 = 118800 X .2613 = 31101.84 cubic inches, the 
answer. 

2. What is the solidity of the middle frustrum of a pro- 
late spheroid, the middle diameter being 60, that of either 
of the two ends 36, and the distance of the ends 80? 

Aris. 177940.224. 



Whence, by substituting this value of a- in the former 
equation, we shall have y'=b--— jj-^ = &^ — 



6V 
5 V— h'^x^ x" 

^ =*— ^( X ^'-^^'') 

And e«mse€[iiently the fluxion of the solid (jyy'^x)=P^^^ — 

^-^ X (b^—K^) ; the fluent of which is = pb'x ~ ^^ X 

pcb^ — pchi 
{¥ — h^); which, when x=c, becomes j)b^c — 

pcX(2h^ + h-) re 



= ^ X 8b^ + 46-. Q. E. D* 



OF ?OWBt» lOl 

S. What is the solidity of the middle frustrum of an ob- 
late spheroid, the middle diameter being- 100, that of either 
of the ends 80, and the distance of the ends 36 1 

Ans, 348814,7g» 

CASE n. 

When the ends are elliptical or perpendicular to the re- 
volving axis* 

RULE.* 

1. Multiply twice the transvei*se diameter of the middle 
section by its conjugate diameter, and to this product add 
the product of the transverse and conjugate diameters of 
either of the ends. 



* Demon. Put so=a, Bo=b, om=zr, on=x, An=y, xc 
= 2, andp=3.14159, &;c. 

Then a i W :-. a^ — x^ : _ X (a^ — x^)=zy^ by the prop- 
a^ 
erty of the ellipsis. 

And, since acd is an ellipsis similar to Emr, it will be 
ry ^ 

b :r :: y : ~jr'=^ j as is shown by. the writers on Conies. 

But the fluxion of the solid aefd is pyzx = pyxX 

ry pry^x prx ¥■ X (a^ — x^) . a^ — x^ 

Y=-j-=-yX- -, :=prbxX—^' And the 

a^ — ^x^ 
^uent=prbxX 1 — • Which, by substituting for a^ its 

6V 2¥-\-y^ ry^ 

value-T2 — 7~2; becomes = prx X — ;jt — = P^ >^ I ^^ + ;jr • 

ry 
And this "again, by putting z for its equal -f", becomes 

=^ X2rb + yz frustrum efda. Or -y^- X (^ef X 2om -f 
adX 2rae)=middle frustrum abgd. Q. E. D* 



15^ 3IBNSURATI0N 

2. Multiply the sum thus found, by the distance of tho 
ends or the height of the frustrum, and the product again 
by .2618, and it will give the solidity required. 

EXAMPLES. 

1. In the middle frustrum ABCD of an oblate spheroid, 
the diameters of the middle section EF are 50 and 30 ; 
those of the end AD 40 and 24; and its height ne 18; 
what is the solidity ? 




Here (50 X 2 X 30 + 40 X 24) X 18 X .261 8= (3000 -f 960) 
X 18 X .2618 1= 3960 X 18 X .2618^= 71280 X .2618 = 
18661. I04i=z solidity required. 

2. In the middle frustrum of a prolate spheroid, the di- 
ameters of the middle section are 100 and 60 ; those of the 
end 80 and 48 ; and the length 36 : what is the solidity ? 

Ans. 149288.832 

3. In the middle frustrum of an oblate spheroid, the di- 
ameters of the middle section are 100 and 60 ; those of the 
end 60 and 36 ; and the length 80 : what is the solidity of 
the frustrum? Ans. 296567.04. 

PROBLEM X\ III. 

Tojind the solidity of the segment of a spheroid. 

CASE I. 
When the hose is parallel to the revolving axis. 



6F SOUDS* 15^ 

RULE * 

1. Divide the square of the revolving- axis by the square 
of the fixed axe, and multiply the quotient by the differ- 
ence between three times the fixed axe and twice the 
height of the seg^ment. 

2. Multiply the product, thus found, by the square of 
the height of the segment, and this product again by ^SSSSj 
and it will give the solidity required* 



1, In the prolate spheroid DEl-'D, the transverse axis 
2 DO is 100, the conjugate AC 60, and the height D» of 
the segment EDF 10 j what is the ^Hdity i 

D 



0Qa 

^ere (j^^x 300-^20) X 10| X ,5236- .86 X 280 x 10« 

X .5236 = 100.80 X 1 00 X .5236 == 10080 X .5236 = 52TT 
S88=s salidity required^ 

2. The axes of a prolate spheroid are 50 and 80 ; what 
is the solidity of that segment whose height is 5, and its 
base perpendicular to the fixed axe ] Ans* 659,736% 



* This rule is formed from the theorem for the segment in th© 
demonstration to problem XX, 

The content of the segment may also be foimd by the following 
theorem: 

(D^4-4(Z^) X -knh = content of the segment; D being the diameter 
of tlie bascj rf= diameter in the middle, As height^ and R5s,7^4sa 
igrea, erf a circle whose diameter is 1, 



154 ME>"SUKATIOX 

3. The diameters of an oblate spheroid are 100 and 60; 
what is the solidity of that segment whose height is 12, and 
its base perpendicular to the conjugate axe ? 

Ans. 32672.64. 

CASE II. 

When the base is perpendicular to the revolving axis. 
RULE.* 

1. Divide the fixed axe by the revolving axe, and mul- 
tiply the quotient by the difference between three times 
the revolving axe and twice the height of the segment. 

2. Multiply the product, thus found, by the square of 
the height of the segment, and this product again by .5236, 
anri \t will give the solidity required. 

EXAMPLES. 

1. In the prolate spheroid aE6F, the transverse axe EF 
is 100, the conjugate ab 60, and the height an of the seg- 
ment a AD 12 ; what is the solidity ? 

* Demon. Put ao=a, i:o=iby om=:r, an=Xy An=:y, nc 
= z, and p= 3.141 6. Then will a^ : 6' : : a'—(a—xy or {2ax 

^ ¥x2ax—z^) 
— x^) : 2 =y^ by the property of the ellipse. 

And since acd is an ellipse similar to ettif, it will be 
ri/ 
b : r :: y : -^=z ; as is shown by the writers on Conies. 

But the fluxion of the solid gacd = pyzx = pyx X 

ry pry'^x prx b^ X (2ax — or) , j^^ 

-17= — ; — =—7: X 2 whose fluent is = — x- 

000a a 

prb 
' — iT~2^' j' which, when a:=A = the height of the segment, 

prb 
becomes (3ah^ — h^)X o~T* ^V hence, since r^a^ we shall 

pb 
have (Sa¥ — h^) X ^ = solidity of the segment. 

Q. E. D. 



OP 


SOLIDS. 

a 






.^- 


"Vh"" 


^ 


sO 








,:-)F 



155 



Here 156 {=.dlf. of Sab and 2an) X If {=EF-^ahx 144 
156X5 
{—square of an) X .5236 = ^ — X 144 X .5236 = 52 X 5 

X 144 X .5236=260 X 144x .5236 = .37440 X .5236 = 
1 9603.584= soZzi^j/?/ required, 

2. Required the content of the segment of a prolate 
spheroid : its height being 6, and the axes 50 and 30. 

Acs. 2450.448. 

PROBLEM XIX. 

To find the solidity of a parabolic conoid* 

RULE.* 

Multiply the area of the base by half the altitude, and 
the product will be the content. 



* Demon* Let Dm=a, B?n=^b, D?2=:ar, iLn=y, and p=: 
3.1416. 

Then, by the nature of the parabola a : b^ :: x : y^, or 

^=zif ; wherefore E — E (=^py^^) = the fluxion of the 
a a 

p¥x" 
solid, and -^5 — = its fluent ; which, when x becomes =a, 

is I" pab- for the whole solid, or for any segment whose 
height is = a, and the radius of its base = b. Q. E. D. 

Coroll. The parabolic coKoid is = I its circumscribing cylinder. 

Note. — The rule given above will hold for any segment of the para- 
boloid, whether the base be perpendicular or oblique to the axe of the 
«olid* 



158 



lONSTTRATlON 



EXAMPLES. 



1. What is the solidity of the paraboloid ADB, whose 
height Dm is 84, and the diameter BA of its circular base 

48? 




^B 



Mere 48^ X .7854 X 42 (=-| Dm) = 2304 x .7854 X 42 = 
1809.5616 X 42 = 76001.5872 = soZi^% required, 

2. What is the solidity of a paraboloid, whose height is 
60, and the diameter of its circular base 100 ? Ans. 235620. 

3. Required the solidity of a paraboloid conoid, whose 
height is 30, and the diameter of its base 40? Ans. 18849.6. 

4. Required the solidity of a paraboloid conoid, whose 
height is 50, and the diameter of its base 100 ? 

Ans. 196350. 



PROBLEM XX. 

ITo find the solidity of the frustrum of a paraboloid^ when 
its ends are perpendicular to the axe of the solid, 

RULE.* 

Multiply the sum of the squares of the diameters of the 
two ends by the height of the frustrum, and the product 
again by .3927, and it will give the solidity. 



* Dfimon. The scirment whose base is B, and altitude A, is 
= J AB, and that whose base is h and altitude ff, is =: J aft, by the 
last problem : wherefore the frustrum, or the difference of the aeg- 



157 



EXAMPLES. 



1. Required the solidity of the parabolic fmstrum ABC<?, 
the diameter AB of the greater end being 58, that of the 
less end dc 30, and the height no 18. 



IZere (58^4- 302)X 18X .3927=(3364 + 900)x 18x .3927 
=4264 X 18 X .3927=76752 X .3927 = 301 40.5104=soKc?- 
ity required* 

2. What is the solidity of the frustrum of a parabolic 
conoid, the diameter of the greater end being 60, that of 
the less end 48, and the distance of the ends 18 ? 

Ans. 41733.0144. 

PROBLEM XXI. 

To find the solidity of cm hyperholoid-. 

RULE.* 

To the square of the radius of the base add the square 
of the middle diameter between the base and the vertex ; 
and this sum multiplied by the altitude, and the product 
again by .5236, will give the solidity. 

Bd 
ment is i ab — ^ah. But b — h : a — a (d) z:b : a= - — 3 ; 

hd 
and B — I d lib : «= • — jhj the uature of the paraboloid; 

and these values of a and a being" substituted for them will 
make i ab — ^ ab=-^^-—^j-=^dX (b-J- J) which is the 
same as the rule. Q. E. D. 

* Demon* Let f= transverse, and c=conjugate cliame- 
ter of the generating hyperbola, 2)=3.1416, ?/, x, the or- 
O 



158 SIENSUEATIOJf 



EXAjMPLES. 



1. In the hyperboloid ACB, the altitude Cr is 10, tlie 
radius Ar of the base 12, and the middle diameter nm 
15.8745 ; what is the solidity 1 

dinates, or semi-diameters of the ends of any frustrurn of 
the hyperboloid, a:=its altitude, and a= distance of the 
less ordinate y from the vertex of the whole solid. 

Then since y^^ ^_Z_-I_-f— ^-i— -^ X c\ we shall have 

At -I- A^ -+- ^AX 4- 

the fluxion of the solid = pY^^ =pc^i X -^ 

tx-\-a^ J •. n * 2 V. At+A^ + Ax + ^tx-\-^.r^ 
— 1— , and Its fluent = pc'xX— — ^ ; 

if At + a'^ Y^ 

and this, by substituting ^ for — , and — 5- ^^^ 

At-\-A?-\-2AX-\-tX^x' , cV 

—^ — — ■ — becomes (^ +2/ — -3^) X W^ — 

solidity of the frustrurn. 

But to convert this into the rules given in the text, let d. 
6, d^ be the greatest, middle, and least diameters, a:= ab- 
scissa whose ordinate is (5, and a= altitude. Then we shall 
have these three equations : 

e^=c^Xt +xXx ^ 

fd^=(^X t -\-x — \a X X — \a 



fiy=c^Xt-\-x^^aXx + ^a 
From the sum of the two latter of which subtract the 
double of the former, and there will result ^Xd^ — 2^-\-d^ 

= iac: and hence -5-,= u • Which bemg 

31' 3 I 

substituted for it in the theorem above will give 

X ap for the content of the frustrum ; which is the same 
as the following rule given in the text. 

And if d the least diameter be supposed to become in- 




Here T5,8745'+122x lOX .5336-251.99975 + 144 X 
10 X .5236 = 395.99975 X 10 X ,5236 — 3959,9975 X 523Q 
==2073.454691 = solidity required. 

2. In an hyperboloid the altitude is 50, the radius of the 
base 52, and the middle diameter 68 ; what is the solidity? 

Ans, 191847, 

PROBLEM XXII. 

To find the solidity of the frustrum of an hyperhoUc eonoidj, 

RULE.* 

Add together the squares of the greatest and least semi- 
diameters, and the square of the whole diameter in the 
middle, then this sum being multiplied by the altitude, and 
the product again by ,5236, will give the solidity, 

^"^^""^"""''^ ~ ^ • jy' + U^ 

finitely little, or nothing, the rule will become — w — ^ 

X a|>=D^+4Q- X a X ,5236, Q. E, D, 

* The demonstration of this rule is contained in that of 
the last problem. 

Or, if D=? middle diameter, m=that at \ of the length, 
g ==generating area of the hyperbola, l == length of the 
spindle, and y:>=^3,1416* 

4?n^'— 3d^ 3s 

Then will (d'-+ -^^^__g^ X — — d) X J-^l — solidity of 

the spindle. And if the generating hyperbola be equilate- 
ral, then will (3s X — — -i,^) X ^pi. ^solidity of the spin* 
die. 



160 



MENSURATION 
EXAMPLES* 



1. In the hyperbolic frustrum ADCB, the length rs is 20, 
the diameter AB of the greater end 82, that DC of the less 
end 24, and the middle diameter nm 28.1708 ; required the 
solidity. 

C 



B 



Here (162-fl2H28.17082) X 20 x .523o9= (2564-144 
+ 793.5939) X 20x .52359 =1193.5939 X 20 X ,52359= 
23S71.878X .52359=::12499.07660202=:5oZic?i72/ required. 

2. Required the solidity of the frustrum of an hyper- 
bolic conoid, the height being 12, the greatest diameter 10, 
the least diameter 6, and the middle diameter 6^. 

Ans. 667.59. 





-^ 


^^ 


4^0 




-^ 


H 


m 






-« 


'^^^ 


^ 


r 


.^^ 



And, if Z=length of the frustrum, s=generating area, 
and the other letters as before; then will (2D* + eP+ 

"4 3 J ^ T"^^ — ^^^12 P^=^^i^i^y of ^^ middle 

frustrum of an hyperbolic spindle. 

But if the generating hyperbola be equilateral, the frus- 
trum will he =(|d?»_Z2+-^ X "^^ )X i;>Z. 

A'bfc. — ^The content of any spindle formed by the ' revolution of a 
conic section about its axis may be found by the following rule : 

Add together the squares of the greatest and least diameters, and 
the square of double the diameter in the middle between the two, and 
this sura multiplied by the length and the product again by .1309 will 
give the solidity. 

And the rule will never deviate much from the truth when the 
figure revolves about any other line which is not the axis. 



OF SOLIDS, ' 161 

3, What is the content of the middle frustrum of an 
hyperbolic spindle, the length being 20, the middle or 
greatest diameter 16, the diameter at each end 12, and the 
diameter at i of the length 14^? Ans. 3248.938- 

4. Required the content of the segment of any spindle, 
its length being 10, the greatest diameter 8, and the middle 
diameter 6. Ans. 272,273. 



Miscellaneous Questions in Solids, 

1, If the diameter of the earth be 7930 miles, and that 
of the moon 2160 miles, required the ratio of their sur- 
faces, and also of their solidities, supposing both of them 
to be globular, as they are very nearly. 

The siu'faces of all similar solids are to each other as 
the squares of their like dimensions ; swch as diameters^ 
circumferences, like linear sides, <^c. <^c. And their solidu 
ties, as the cubes of th^se dimensions,^ 

Hence the surface of the ?noon : surface of the earth : ; 
. ■ 2160^ 4665600 1 

2160-7930^ and* ^gs¥ = 62HSj900=-l3A7 «^' ^^ ^ = 
13| nearly. 

Also the solidity of the moon : solidity of the earth : : 
2160=^ 1 
2160V: 7930=^ and Zg^s^-j^ nearly, or, As I : 49| 

Ans. 

2. Three persons having bought a sugar-loaf, want to 
divide it equally among them by sections parallel to the 
base; it is required to find the altitude of each person's 
share, supposing the loaf to be a cons \vht>se height is 20 
inches. 



* The ratio of one quantity to another may be expresseo by dlvid. 
mg ibe antecedent by the consequent 
02 



162 



MEKSXrSATION 



20» 



By the similar cones we have* 3:1:: 20* : -^ = the 

o 

cube of the height of the top sections,' wherefore s/-^ = 
13»86T the upper part. 

2X20* , . 2x20=» 



Also 3 : 2 : : 20* : - 



and s/ 



13.867 = 



3 "^"^ ^ 3 
3.604 the middle part; wherefore the lower part will he 
2.528. 

3. Three men bought a tapering- piece of timber, which 
was the frustrum of a square pyramid ; one side of the 
greater end was 3 feet, one side of the less end 1 foot, and 
the length 18 feet ; what is the length of each man's piece, 
supposing they paid equally, and are to have equal shares? 

E 



1 


Ac 


t/ oi 


i \ 


TiyJ PI 


' \]y3 


/ 1 


1 \ 



F G B 



* This proportion as well ^s all others of the kind, may be ex- 

20 

pressed thus : ^ 3 : v^ 1 : : 20 : — — =the height of the top section ; 

V 3 
andj in seme instances, this is the more convenient method. 



OF sotii»s» 168 

Let ABCDE he a section of the pyramid (when com* 
pleted) passing through the vertex^ and bisecting the oppO' 
site sides of tJie hasey and let IL and MN represent the re^ 
quired sections. Draw EF to the middle of AB, and draw 
CG parallel to it* 

Then by similar triangles BG (Ifoot) : GC (18) : : BF 
(1,5) : FE(27) and FE— -FH=EH=9, the altitude of the 
pyramid EDC« 

Hence Prob. VTIL of solids the solidities of the ttco 
pyramids EAB and EDC ivill be found 81 and 3 cubic feet 
respectively, and 81 — 3=78=^Ae solidity of the frustrum 

78 
ABCD. Also —=26, the solidity of each person's share^ 

which added to the solidity of EDC, tvill give the solidity 
of EIL=29, and the double of it added to EDC will give 
the solidity of EMN^dS, 

Now in the similar pyramids, EDC(3) : EIL(29) : : EH' 
: EO*, the cube root of which will give EO and EO — ^EH 
==HO the length of the part adjacent to the less €nd= 
10.172. 

Again EIL(29) : EMN(55) : : EO* : EF the cube root of 
which will give EP and EP — EO=OP the length of the 
middle part= 4.559. Lastly, EF — ^EP=:PF the length of 
the part adjacent to the larger endz=S.269. 

4. If a round pillar, 7 inches over, have 4 feet of stone 
in it ; of what diameter is the column, of equal length, that 
contains 10 times as much? 

The solidities of cylinders, prisms, parallelopipedons, 
<^c. which have their altitudes equal, are to each other as 
the squares of their diameters or like sides. The same re- 
mark is applicable tofrustmims of a cone or pyramid when 
the altitude is the same, and the ends proportional. 

Hence, As 4 : 40, or As 1 : 10 : : 7^ : 490=^^.6 square 
of the required dmrneter^and V490=22.1359 the diameter 
required. 

5. There is a mill-hopper, in the form of a square pyra- 
mid, whose solid content is 13^ feet ; but one foot is cut 



164 MENSURATION OF BOUDt. 

off its perpendicular altitude to make a passage for the 
grain, from the frustrum or hopper to the mill-stone ; the 
sides of its-greater and less end are in proportion of 4^ to 
1. Required the content in dry or corn measure. 

Ans. 10.7292 bushels, 

6. The ditch of a fortification is 1000 feet long, 9 feet 
deep, 20 feet broad at bottom, and 22 at top ; how much 
water will fill the ditch, allowing 282 cubic inches to make 
a gallon] Ans. 1158127f| gallons, 

7. A person having a frustrum of a cone 12 inches in 
height, and the diameters of the greater and smaller ends 
5 and 3 inches respectively, wishes to know the diameter 
of a frustrum of the same altitude, that would contain 3666 
cubic inclies, and have its diameters in the same proportion 
as the smaller one. 

Ans. The greater diameter 24.4002, and less 14.6401. 



REGULAR BODIES, 



A REGULAR BOBY IS E soHd Contained under a certain 
number of similar and equal plane figures. 

The whole nimiber of regular bodies which can possibly 
be formed is five, 

1. The Tetraedron^ or regular pyramid, which has four 
triangular faces. 

2. The Hexaedron, or cube, which has six square faces. 

3. The Octaedron^ which has eight triangular faces, 

4. The Dodecaedron^ which has twelve pentagonal facc§» 

5. The Icosaedron, which has twenty trimigular faces. 
If the following figures are made of pasteboard, and the 

lines be cut half through, so that the parts may be turned 
up and glued together, they will represent the five regular 
bodies here mentioned. 




166 



SKaX7X4AK COPIES, 




PROBLEM I. 

To find the solidity of a tetraedron* 

RULE,* 

Multiply J^ of the cube of the linear side by the square 
root of 2j and the product will be the solidity. 



* Demon, From one angle c of the tetraedron abc?^ 
let fall the perpendicular ce, upon the opposite side, and 
draw AC. 

Then ac^ — Ae^=ce^; and since the point c is equally 
distant from the three angles a, b, and ??, ^ ac'= (^ ab^) 
A.e^ as is shown in the demonstration of the rule for regu- 
lar polygons, page 61. Consequently ac^ — ^AC^=fAC* 
= ce^ or ce = ac v/ §• But the area of the triangle 
AnB=:|AB^v^3=:^ ac^a/S; and therefore i AC V f (^ ce) 
X iAC^ v/ 3( AAr?B) = JjAC^ v/ 2. Q. E. D. 

If L be put = length of the linear edge, then will l^>/3 
= whole surface of the tetraedron. 

The rule for the hexaedron, or cube, has been given be- 
fore. 



"REGTrLAJS. BODIES. 167 



1. The linear side of a tetraedron ABCn is 4: what is 
the solidity 1 




A 



e 4X4X4 4X4 16 16 

12^ n/2=-Y2— X V2=-3-X v^2=^X v/2=y X 

22.624 
1.414^ — ix — z—7 .5AlS^:^sohdit7/ required. 

2* Required the solidity of a tetraedron whose side is 6. 

Ans. 25.452. 

PROBLEM n. 

Tojind the solidity of an octaedron* 

RULE.* 

Multiply i of the cube of the linear side by the square 
root of 2, and the product will be the solidity. 



* Demon. From the angle d of the octaedron dbga let 
fall the perpendicular dc. 

Then since the solid is composed of two equal square 
pyramids, each of whose bases buac are equal to the square 
of the linear side ag or ad, we shall have buacX f De=A7i* 
>: |De= content of the solid. 

But De evidently bisects the diagonal ba, and is equal to 



169 



SECULAR BODIES. 



1. What is the solidity of the octaedron BGAD, whose 
linear side is 4? 




4» 64 

— Xv^2=g-X v/ 2 = 21.333, &c. X v/ 2=21.333, &c. 

X 1.414, ^c. — S0.l6^S6=z solidity required. 

3» Required the solidity of an octaedron whose side is 8. 

Ans. 241.3568, 

PROBLEM im 

To^find the solidity of a dodecaedron* 

RtFLE.* 

To 21 times the square root of 5 add 47, and divide the 
sum by 40 : then the square root of the quotient being 



Be ; therefore aw^X foe = ah^X |bc=aw- X ^ba = ^aw^X 
>/Aw2 + Ac^=iA/iV2An2=^A7iV2. Q. E. D. 

If L = linear side as before, then will 2l^v^ 3= surface 
of the octaedron. 

* Xhmeris Let a be a solid angle of the dodecaedron, and ac a 



REGULAR BODIES. 169 

multiplied by 5 times the cube of the linear side will give 
the solidity required. 



perpendiciilar falling on the equilateral 
plane BDF. Also join tlie points D, c and 
c, F. 

Then the angle DcrF contains 108 de- 
grees, whose sine is | v'lO-f 2^/5, and 
ike angle aFD contains 36 degrees, whose 
sine is ?>/ 10 — 2v/o, the radius in both 
cases being taken equal to 1. 




Therefore, by trigonometry, iv/10 — 2^/5:^^10 + 2^/6 
5+v/5 I+n/5 

:: aD : Dr=aDv/ ^ ^ =:nD — ^ — =i + iV5XaD. 

Again, since c is the centre of DBF, the angles cDF and cFD are 

'each 30°, and the angle DcF=120°; but the suie of 30° is J; and 

the sine of 120° is i v^3 ; whence, by trigonometry, ^ y/3 : df : : ^ 

DF 1 + n/5 . 

: D6*=-To=c^D 2^3 ' ^^^ consequently ac= y/ od^—dc^ 
3— V5 



—aDs/ — Q — =aT)s/i — iv/5. 

But a perpendicular from a upon the plane BDF must pass through 
the centre of the circumscribing sphere, and ac will be the versed 
sine of an arc whose chord is aD, and radius equal to that of the said 
sphere. 

„,, -r^ ^^' ^^^ N/3+V15 

Whence ac : aD : : a : — = , , — ;-=7^=aD — - — 

= diameter of the circumscribing sphere, and aD — 

=R= radius of the circumscribing sphere. 
Again, the angle ¥on contains 72°, whose sine is i 

V^104-2v/5;and the angle oFn is 54°, whose sine isllL^ . 

4 ' 

. l + v/5 

whence by trigonometry, \'s/lQ-\-2s/5'. — -7 • • ^^ («») 

I + n/5 5+^5 



170 EEGULAR BODIES. 

EXA3IPLES. 

1. The linear side of the dodccaedron ABCDE is 3 
what is the solidity ? 

D 




21-75 + 47 21x2.23606 + 47 
V — -^ X 27 X 5 = V 1^ X27X5= 

46.95726 + 47 
V V7J X 135=206.901 solidity required, 

2. The linear side of a dodecaedron is 1 ; what is the so- 
lidity? Ans. 7.6631. 

But since the radius of the circumscribing sphere is the hypothe- 
nuse of a right angled triangle, whose legs are oF and the radius 

of the inscribed sphere, we shall have \/r^ •— op^ = 

25 + llv/5 



v/(ix/3+ VlSao)^ — ^ +y'ov/5aD^ = av sf - 



40 



\/f + |^^V 5= radius of the inscribed sphere. 

And because the solid is composed of 12 equal pentagonal pyramids, 

5aD^ 

each of whose bases are by Prob. VUI. = -— p V 1 + 1 V 5 ; 

eOoD^ eOr/D^ OD 

therefore -j- Vl+f v/5x ir = ^ Vl + fv/SX y 

25 + 11^5 ^ 47 + 21 v/5 
\f Tq =5aD^ V jTj = solidity of the dode- 

caedron. Q. E. D. 

If L be put for the linear side, then will Ibi}-^ = — 

o 

=surface of the dodecaedron. 



SE6FI.AH BODIES. 



ni 



PROBLEM IV. 
To find the solidity of an icosaedron, 

RULE.* 

To 3 times the square root of 5 add 7, and divide the 
sum by 2 ; then the square root of this quotient being mul- 
tiplied by I of the cube of the linear side will give the so- 
lidity required. 

■;7 + 3v/5 

That is f S'X s/K ^ )== solidity when S is=to the 

linear side. 



* Demon. Let A be a solid angle of the 
icosaedron, formed by 5 faces, or triangles, 
whose bases make the pentagon BCDEm. 

Then, if a perpendicular be demitted from 
A upon the pentagonal plane BCDEm, it will 
fall into the centre n, and Bn, by the demon- 
Etration of the last problem, will be = ab 

^/ , and the radius of the circle 

circumscribing one of the faces ABC = Jab -«/ 3. q 

But the radius of the circumscribing sphere is R= 

5+ a/5 




2xn 



~2 J A^^—-&ir ~^^^ ~"~8 ' ^^^^^ ^^ ^^ ^^® ^^^^ problem. 

And, since R is the hypothenuse of a right angled triangle, one of 
whose legs is J ab ^3, the radius of the circle circumscribing the 
face ABC, and the other r, the radius of the inscribed sphere, we shall 

5+\/5 

: y/ ^- AB^ iAB^ == AB 



have r = Vr'— Kab V 3)' 

7 + 3^5 
^ 24 . 

But the solid is composed of 20 equal triangular pyramids, each of 
AB^ 20ab3 

whose bases is=^x/3 by Problem VIII. ,- therefore —j- 



178 



REGUl^AR BODIEi. 



1. The linear side of the icosaedron ABCDEF is 3j 
what is the solidity ] 

D 




3^5 + 7 5XS^ 3x2.2360 6 + 7 5x27 



AB 7 + 3v/5 , 7 + 3v/5 
v/3xi r = 5abV3x yV — 24 =6 abV — ^ = 

•solidity of the icosaedron. Q. E. D. 

If L be put for the linear side, tlien will SL^ ^Z 3 = surface of the 
icosaedron. 

Note. — The superficies and solidity of any of the five regular bodies 
may be found as follows. 

Rule 1. Multiply the tabular area by the square of the linear edge, 
and the product will be the superficies. 

2. Multiply the tabular solidity by the cube of the linear edge, and 
the product will be the solidity. 

Surfaces and Solidities of the Regular Bodies. 



No. of 
sides. 

4 

6 

8 
12 
20 


Names. 


Surfaces, 


Solidities. 


Tetraedron 

Hexaedron 

Octaedron 

Dodecaedron 

Icosaedron 


1.73205 
6.00000 
3.46410 
20.64.578 
8.66025 


0.11785 
1.00000 
0.47140 
7.66312 
2.18169 



KEGULAR BODIES. 173 

' 6.70818-1-7 5X9 13.70818 45 . 

y/ 2 X-2-=V 2 ^ y =v/ 6.85409 X 

22.5=2.61803 X 22.5=58.9056=5oZi£Z% required, 

2. Required the solidity of an icosaedron, whose linear 
side is 1 1 Ans. 2.1817. 



P3 



CYLINDRIC RINGS. 



PROBLEM I. 

To Jind the convex superficies of a cylindric ring* 

RULE.* 

To the thickness of the ring add the inner diameter, and 
this sum being multiplied by the thickness, and the product 
again by 9.8696, will give the superficies required. 

EXAMPLES. 

1. The thickness of Ac of a cylindric ring is 3 inches, 
and the inner diameter cd 12 inches ; what is the convex 
superficies ? 




* A solid rin;^ of this kind is only a bent cylinder, and therefore 
the rules for obtaining its superficies, or solidity, arc the same as those 
already given. For let Ac be any section of the solid perpendicular to 
its axis on, and then Ac X 3.1 41 5 J, &.c. = circumference of that section, 
and A< -\-cd (oh) X 3.14159, «SL.c.=lengtli of tJie axis on. 



OYUNDRIO RINGS. 175 

12 + 3 X 3 X 9.8696 = 15x 3x 9.8696 = 45x 9.8696 = 
444. 132= superficies required* 

2. The thickness of a cylindric ring is 4 inches, and 
the inner diameter 18 ; what is the convex superficies? 

Ans. 868.52 square inches* 

3. The thickness of a cylindric ring is 2 inches, and the 
inner diameter 18 j what is the convex superficies? 

Ans. 394,784 square inches* 

PROBLEM n. 

To find the solidity of a cylindric ring* 

RULE.* 

To the thickness of the ring add the inner diameter, and 
this sum being multiplied by the square of half the thick- 
ness, and the product again by 9.8696, will give the solidity. 



Whence acX 3.14159, &c.Xac + cc^x 3.14159, &;c. = 
Ac+cc?X AC X 3.14159, &;c. p=AC + ccZx acX 9.8696, &c* 
= superficies required. 

3.14159 
* Demon, ac'x .78539, &c.=ac2x -7— , &c. = i 



Ac^X 3.14159, &c. = area of the section ac; and Ac-\-cd 
(on)X 3.14159, &;c.=Iength of the axis on. 

Therefore Ac^cd X ^ac^ X 3.14159, &c. 'f = Ac + cdxi 
Ac^x 9.8696. Q. E. D. 

This figure being- only a cylinder bent round into a ring-, its surface 
and solidity may also be found as in the cylinder, namely, by multi- 
plying the axis or length of the cylinder by the circuniferenco of the 
ring, or section, for the surface, and by the area of a section for the 
solidity. 

Thus, if c= circumference of the ring, or section, c=area of that 
section, and Z^=lcngth of the axis ; then will cZ=surface of the ring, 
and aZ=::to its solidity-. Which rules are the same as for tlie cylinder, 
and may be easily converted into those given in the text. 

These rules arc indeed so obvious, as to render any demonstration 
of them altogether imnecessary. 



176 CYLINDRIC RINGS. 

EXAMPLES. 

1. What is the solidity of an anchor ring, whose inner 
diameter is 8 inches, and thickness in metal 3 inches ? 

8T3x"||2 X 9.8696 = 11 X 1.52 x 9.8696=11 X 2.25 X 
9.8696=24.75 X 9.8696 = 244.2726=soZiJi<y required, 

2. The inner diameter of a cylindric ring is 18 inches, 
and its thickness 4 inches ; what is the solidity ? 

Ans. 868.5248. 

3. Required the solidity of a cylindric ring, whose thick- 
ness is 2 inches, and its inner diameter 12. 

Ans. 138.1744. 

4. What is the solidity of a cylindric ring, whose thick- 
ness is 4 inches, and inner diameter 16? Ans. 789.568. 



ARTIFICERS' WORK. 



Artificers estimate or compute the value of their 
works by diflferent measures, viz,* 

1. Glazing and Mason'' s flat work, &;c. by the foot. 

2. Painting, Plastering, Paving, &lc. by the yard. 

3. Flooring, Partitioning, Roofing, Tiling, &c. by the 
square of 100 feet. 

4. BricJcworJCi &c. by the rod of 16^ feet, whose square 
is 272^, 



The measures made use of in these works are contained 
in the following table : 



12 inches 
144 square inches 

9 square feet 
100 square feet 
272^ square feet, or > 

30^ square yards, ) 



make 



1 lineal foot. 
1 square foot. 
1 square yard, 
a square. 

1 rod, perch, or square 
pole. 



* The best method of taking the dimensions of all sorts of artificers* 
work is by feet, tenths, and himdredths ; because the computation may 
then be performed by common multiplication, or by the sliding- rule, 
hereafter described. 



BRICKLAYERS' WORK.* 



Bricklayers compute or value their work at the rate of 
a brick and a half thick, and, if the wall be more or less 
than this standard, it must be reduced to it as follows : 



* Note. — In practice it is usual to divide the square feet by 272 
only, omitting the | ; but the more accurate way is, to divide by 
272^5. 

The usual way to take the dimensions of a building is to measure 
half round its middle, on tlie outside, and half round it on the inside ; 
and this will give the true compass, in wliich the thickness of the wall 
is included. 

When the height of the building is unequal, take several different 
altitudes, and their sum, being divided by the nmnber of zdtitudes you 
have taken, may be considered as the mean height. 

To measure a chimney standing by itself, without any party-wall 
adjoining ; girth it about for the length, and reckon the height of the 
Btory for the breadth ; but if it stand against a wall, you must measure 
it round to the wall for the girth, and take the height as before. 

When the cliimney is wrought upright from the mantel-tree to the 
ceiling, the thickness must always be the same \\ith tlie jambs ; and 
nothing is ever deducted for the vacancy between the floor and the 
majitel-tree, because of the gatliering of the breast and wings to make 
room for the hearth in the next story. 

To measure chimney shafts, or that part which appears above the 
roof; girth them with a line, about the least place for the length, and 
take the height for the breadth ; and if they be four inches thick, set 
down the thickness at one brickwork ; but if they are 9 inches thick, 
reckon it a brick and a half, in consideration of the plastering and 
BcafTolding. 

All windows, doors, &.c. are to be deducted out of the contents of the 
walls in which they are placed. But this deduction is made only with 
regard to materials ; for the value of their workmanship is added to 
tlie bill at tlie stated rate agreed on. 



bricklayers' work. 179 



RULE. 

Multiply the superficial content of the wall in feet, by 
the number of half bricks in the thickness, and i of that 
product will be the content required. 

Note. — In America, bricklayers' work is generally reck- 
oned by the 1000. 



1.* How many square rods are there in a wall 52^ feet 
long, 12 feet 9 inches high, and 2^ bricks thick ] 



By Decimals. 



Here 5^.5 X 12.75 _669j75 ^ ^^^^^^ 
272 ~ 272 



2.4609 

5 half bricks. 



3)12.3045 

ro. ft. in, 

4.1015=4 27 7 answer. 



There are also other allowances to be made to tlie workmen, such 
as those for returns or angles made by two adjoining walls, and double 
measure for feathered gable-ends, &c. 

All ornamental work is generally valued by the foot square, such 
as arches, doors, architraves, friezes, cornices, &c. But carved 
mouldings, &c. are often agreed for hj the running fix)t, or lineal 
measure. 

* In this and the following examples, 272 feet are used for a rod. 



180 bricklayers' work. 

By Cross Multiplication* 



feet in, 
52 6 
12 9 




630 
39 4 6 




272)669 4 6 




2 124 4 
5 

8)12 82 8 

^ 21 1 as 


before 



2. How many square rods are there in a wall 62^ feet 
long, 14 feet 8 inches high, and 2^ bricks thick? 

ro.fe. in. p. 
Ans. 5 167 9 4 

3. If each side wall of a building be 45 feet long on the 
outside each end wall 15 feet broad on the inside, the 
height of the building 20 feet, and the gable at each end 
of the wall 6 feet high, the whole being 2 bricks thick ; 
what is the true content in standard rods T 

Ans. 12.2059 



MASONS' WORK. 



To Masonry belong all sorts of stone work, and the 
measure made use of is a solid perch, or a superficial or 
solid foot. 



EXAMPLES. 

1. Required the solid content of a wall whose length is 
48 feet 6 inches, its height 10 feet 9 inches, and thickness 
2 feet. 

By Decimals* 
48.5xl0.T5x2=1042.75/ee^ An§. 



* Solid measure is principally used ibr materials, and the superfi- 
cial for workmansliip. — In the solid measure, the true length, breadth, 
and thickness, are taken, and multiplied continually together. And 
in the superficial measure, the length and breadth of every part of the 
projection must be taken, as it appears without the general upright 
face of the building. 

Masons, in measuring their work, usually take the whole girth of the 
building, tliat is, the length of a string that passes entirely round the 
building, which is 4 times the thickness of the wail more than the 
true meiisure. This is added on account of the trouble of carrying 
up the corners. 

In America, the thickness of the wall is not reckoned to the mason 
at less than 18 inches ; but if it is more than that thickness, it is re- 
duced to it. No dsduction of ths mason work is made for doors, 
windows, «fcc. on account of tlie trouble of carrying up the straight 
walls on the sides of them. 

All windows, doors, &,c. are to be deducted cut of the contents c?f 
the walls in which they are placed with regard to materials. 

Q 



182 MASONS* WOSK. 

By Cross Multiplicadofu 

feet in. 
48 6 
10 9 



485 







36 


4 


6 


521 


4 


6 


2 









1042 9 the same as before, 

2. Required the solid content of a wall whose length is 
53 feet 6 inches, its height 12 feet 3 inches, and its thick- 
ness 2 feet. Ans. 1310yce^ 9 in. 

3. What is a marble slab worth, whose length is 5 feet 
7 inches, and breadth 1 foot 1 inches, at 80 cents per foot ? 

dolls, cts, 7n> 

Ans. 8 18 8. 

4. What is the solid content of a wall, whose length is 
60 feet 9 inches, its height 10 feet 3- inches, and its thick- 
ness 2^ feet? Ans. 1556.71 875/ee^ 

5. In a chimney-piece, suppose the fe. in. 
Length of the mantel and slab, each ... 46"^ 

Breadth of both together, 3 2 I 

Length of each jamb, - - 4 4 j 

Breadth of both together, 1 9 J 

What will be the content of the chimney-piece ? 

Ans. 21 feet \0 in, 

6. The dimensions of a certain building are as follow : 
viz. 58 feet by 26 on the outside; height of the building 
22 feet, height of the gable at each end 12 feet, thickness 
of the wall 15 inches. Two doors 4.j by 8 feet each, 28 
windows each 3| by 6 feet: what will the mason work 
amount to at 56 cents a perch, and what must be paid for 
the stone at 44 cents a perch T 

An3. For the work, $136 03 ; For the stone, $71 95^ 



CARPENTERS' AND JOINERS' AVORK.* 



Carpenters' and Joiners' work is that of flooring, par« 
litioning, roofing, &c, and is measured by the square of 
100 feet. 



* JVbie.— -Large and plain ajticles are usually measured by the foot, 
or yard, &c, square ; but enriched mouiduigs, and some other articles, 
are often estimated by running or Uneal measure, and SQxne things are 
rated by the piece. 

In measuring of joists it is to be otserved, that only one of their 
dimensions is the same with that of the floor, and the other will ex, 
eeed the length of tlie room by the thiclaiess of the wall, and cne^ 
third of the same, because each end is let into the wall about two^ 
thirds of its thickness, 

No' deductions are made for hearths, on aceoimt of the additional 
trouble and waste of materials, 

Partitians are measured from wall to wall for one dimension, and 
from floor to floor, as far as they extend, for the other. 

No deduction is made for door-ways, on account of the trouble 
of framing them, 

In measuring of Joiners' work, the string is made to ply close to 
every part of the work over whic/i it passes. 

In TGof.ng, the length of the house in the inside^ togetlier with tAVo- 
thirds of the thicliiiess of one gable, is to be considered asthe lengthy 
and the breadtli is equal to double the lengtli of a string, which is 
stretched from the ridge dov\^n to tlie rafter, along the eaves^board, 
till it meets with the top of the wall. 

For stair-cases, take the breadth of an the steps, and make a line 
ply close over them, from the top to the bottom, and multiply the 
length of this line by the length of a step for the whole area, By the 
length of a step is meant the length of the front and the returns at 
the two ends, and by the breadth is to be imderstood the girth of ks 
two upper surfaces, or the tread and riser. 



184 CASPENTERS* AND JOINERS* WORK* 



1. If a floor be 57 feet 3 inches long, and 28 feet 6 
inches broad, how many squares will it contain? 

By Decimals* 

57.25 X 28.5=1631,625 sq.ft.— \Q sq. Sift, 7 m. 6". Ans. 

By Cross Midtiplication* 



feet 
57 

28 


in* 
3 
6 




1603 

28 



7 


6 


16,31 


7 


6 



For the hahistradey take the whole length of the upper part of the 
haud-rail, and girth over its end till it meet the top of the newel-post, 
for the length ; and twice the length of tlie baluster upon the landing, 
with the girth of the hand-rail, for the breadth. 

For wainscoting, take the compass of the room for the length, and 
the heif'-ht from the floor to the ceiling, making the string ply close 
into all the mouldings, for the breadth. Out of this must be made 
deductions for windows, doors, chimneys, &c. but workmanship ia 
covmted for the whole, on account of the extraordinary trouble. 

For doors^ it is usual to allow for their thickness, by adding it into 
both the dimensions of length and breadth, and then multiplying them 
together for tlie area. If the door be panelled on both sides, take 
double its measure for the workmanship; but if one side only be 
panelled, take the area and its half for the workmanship. 

For the surrounding architravCy gird it about the outermost part for 
its length ; and measure over it as far as it can be seen when tJie door 
is open, for the breadth. 

Window-shutters, bases, &-c. are measured in the same manner. 

In the measuring of roofing for workmanship alone, all holes for 
chimneys-shafts and sky-lights are generally deducted. 

But in measuring for work and materials, they commonly measure 
in all sky-Iights, luthern-lights, and holes for the chimney-shafts, on 
account of their trouble and waste of materials. 



carpenteks* and joiners' work. 185 

13, A floor is 53 feet 6 inches long, and 47 feet 9 inches 
broad ; how many squares does it contain ? 

Ans. 25 sq. and b^feet 1\ in, 

3. A partition is 91 feet 9 inches long, and 1 1 feet 3 
inches broad : how many squares does it contain 1 

Ans. 10 5^'. and S2feet, 

4. *If a house within the walls be 44 feet 6 inches long, 
and 18 feet 3 inches broad; how many squares of roofing 
will cover it, allowing the roof to be of a true pitch 1 

Ans. 12 sq. and IS feet. 

5. If a house measure within the walls 52 feet 8 inches 
in length, and 30 feet 6 inches in breadth, and the roof be 
of a true pitch, what will it cost roofing, at 1 dollar 40 cts, 
per square 1 Ans. 33 dolls. 73 cts. 3 tr. 

* It is customary to reckon the flat and half of £iny building within 
the walls, for the measure of the roof of that building, when the roof 
is of a true pitch, or so that the length of the rafters is | of the breadth 
of the building. 



Q8 



SLATERS* AND TILERS' WORK. 



In these works the content of a roof is found by multi- 
plying the length of a ridge by the girth from eave to eave ; 
and, in slating, allowance must be made for the double row 
at the bottom. 

In taking the girth, the line is made to ply over the 
lowest row of slates, and returned up the under side till it 
meet with the wall or eaves-board; but in tiling, the line 
is stretched down only to the lowest part, without returning 
it up again.' 

Double measure is generally allowed for hips, valleys, 
gutters, &c. but no deductions are made for chimneys.* 



* In angles formed in a roof^ running from the ridge to the eaves, 
that angle of the roof which bends inwards, is called a ralley : and 
the angle bending outwards is called a hip. And in tiling and slating, 
it is comiron to add the lengtli of the valleys to the content in feet; 
and sometimes also the liips arc added. 

In slating it is common to reckon the breadth of the roof 2 or 3 in- 
ches broader than what it measures, because the first row is almost 
covered by the second ; and this is done somctinies when a roof is 
tiled. 

Note. — Sky-lights and cliimncy-shafts arc always deducted; but 
they seldom deduct lutliem-lights, or garret-windows on the roof; for 
the covering them is recloned equal to the hole in the roo'". 

In all works of this kind the content is computed, eitlier. in yards 
of 9 square feet, or in squires o^ a hundred iect, and the same allow- 
ance of hips and valleys is to be made as in roofing. 



fiLATEBs' AND TILERS* WORK. 187 



1. The length of a slated roof is 45 feet 9 inches, and 
its girth 34 feet 3 inches ; what is the content ? 

By Decimals^ 

45.75X 34.25=1566.9375 s^. ^.=174.104 yds, Ans. 

By Cross Multiplication, 

feet irin 
45 9 
34 3 



1555 6 
11 5 3 

9)1566 11 3 

174 11 3 

Ans. 174 yds, 

2, What will the tiling of a barn cost at $S 40 per 
square, the length being 43 feet 10 inches, and the breadth 
27 feet 5 inches, on the flat, the eave -boards projecting 16 
inches on each side, allowing the roof to be of a true pitch ? 

Ans. 865 26. 



PLASTERERS' WORK. 



Plasterers' work is of two kinds, viz. plastering upon 
laths, called ceiling, and plastering upon walls, called ren- 
dering; and these different kinds must be measured sepa- 
pately, and their contents collected into one sum.* 

Note — Proper deductions must be made for doors, win- 
dows, &;c. 

EXAMPLES. 

If a ceiling be 59 feet 9 inches long, and 24 feet 6 in- 
ches broad ; how many yards does it contain ? 

By Decimals. 
59.75X24.5=1463.875 5^. /ee^= 162.6528 sj- V^s. Ans. 
By Cross Multiplication* 
feet in. 



24 6 




1434 
29 10 


6 


9)1463 10 
162 5 10 


6 

6 
Ans. 162 yards bfeet. 



* Plasterers' plain work is measured by the square foot, or yard of 
9 square feet; and enriched mouldings, &e. by running or lineal 
measure. 



PLASTERERS* WORK. 189 

8. If the partitions between rooms are 141 feet 6 inches 
about, and 1 1 feet 3 inches high j how many yards do they 
contain? Ans, 176.87. 

3. The length of a room is 14 {eet 5 inches, its breadth 
13 feet 2 inches, and height 9 feet 3 inches to the under 
side of the cornice, whose girth is 8^ inches, and its pro- 
jection 5 inches from the wall on the upper part next the 
ceiling ; what will be the quantity of plastering, supposing 
there -be no deductions but for one door, whose size is 7 
feet by 4 feet ? 

{ 53 yds. oft. 3 in. of rendering* 
Ans. < 18 yds. 5 ft. 6 in. of ceiling. 
( And 39 ft. 0|i in. of cornice* 



PAINTERS' AISTD GLAZIERS' WORK. 



Painters* work is measured in the same manner as that 
of Carpenters ; and in taking the dimensions, the line must 
be forced into all the mouldings and corners. 

The work is estimated at so much per yard, except 
sashes, which are done at so much per light ; in carved 
mouldings, &c. it is customary to allow double the usual 
measure. 

Glaziers' work is done at so much per light, that is, com- 
puted at a given price for putting in each pane of glass, ac- 
cording to the size. 

EXAMPLES. 

1. If a room be painted, whose height is 16 feet fl 
inches, and its compass 97 feet 9 inches j how many yards 
does it contain ? 



Balustrades and most other works of tliat kind are measured as in 
Joiners' work, excepting for doors, window-shutters, &c. where tho 
Joiner is only allowed the area and half area ; but the Painter haa 
always double the area of one side, because Qyerj part tliat is painted 
must be measured. 

Note, — Painters take their dimensions with a string-, and measure 
from the top of the cornice to the floor, girthing tlie string over all the 
mouldings and swellings ; and their price is generally proportioned to 
the nmnber of times they lay on their color. 

All work of this kind is done by the square yard, and every part 
where the color lies must be measured and estimated in the general 
account of tlie work. 

Deductions are to be made for chimneys, casements, &c. ajid tlie 
price is generally proportioned to the number of times Uiey lay on 
tlieir color. 



painters' and glazier?' work. 191 

By Decimals* 
97.75 X 16.5=1612.875 sq.feet=179.208sq. yds. Answer. 
By Cross Multiplication* 
97 9 



16 


6 




1564 

48 



10 


6 


9)1612 


10 


6 



179 1 10 6 

Ans. 179 yards Ifoot 

2. The height of a room is 14 feet 10 inches, and the 
circumference 21 feet 8 inches: how many square yards 
does it contain? Ans. 35.71. 

3. Suppose a room, that was to be painted at Sd, per 
yard, measures as follows : the height is 11 feet 7 inches; 
the girth or compass 74 feet 10 inches ; the door 7 feet 6 
inches, by 3 feet 9 inches ; five window-shutters, each 6 
feet 8 inches, by 3 feet 4 inches ; the breaks in the win- 
dows 14 inches deep and 8 feet high ; the chimney 6 feet 
9 inches by 5 feet ; a closet, the height of the room, 3^ 
feet deep, and 4|: feet in front, with shelving, at 22 feet 6 
inches by 10 inches; the shutters, doors and shelves being 
all colored on both sides ; what will the whole come to ? 

Ans. 4Z. 185. 9d, 



PATIORS' WORK. 



Paviors' work is done by the square yard, and the con- 
tent is found by multiplying the length by the breadth. 

Or if the dimensions be taken in feet, and the area be 
found in the same measure, the result being divided by 9 
will give the number of square yards required. 



EXAMPLES. 

1. What* will the paving of a rectangular court-yard 
come to at 3s. 2d, per yard, supposing the length to be 27 
feet 10 inches, and the breadth 14 feet 9 inches? 



Plumbers' work is gfenerally done by tlic pound, or liundrcd weight, 
and the price is regulated according to the vakie of the lead at the 
time the contract is made, or wlien the work is performed. 

Sheet lead, used in roofing, guttering, &c. is generally between 7 
and 121bs. weight to the square foot 

The following table will show the weight of a square foot to each 
of these thicknesses. 



Thick. 


/6s. 
sq. foot. 


Tldck. 


Lbs. 

sq. foot. 


Thick. 


lbs. 
sq.fnot. 


.18 
.14 


7.373 

7.668 
8.258 
8.427 


.15 
.16 

.17 


8.848 

9.438 

9.831 

10.028 


.18 
.19 

A 


10.618 
11.207 
11.797 
12.387 



PAVIOKS' WORK. 



193 



By Cross Multiplication. 

feet in. 
27 10 
14 9 



389 8 
20 10 6 

9)410 6 6 

45 5 6 6 a/ 3s. 2d. 





9. 


d. 




2 







1 



2 


/. 


In. 


V- 


3 








1 








1 











6 











6 



45 


4 10 




2 5 




7 


6 


1 


Oi 





4 





4 





2 








7 4 


4il 



4^ the answer. 

2. A rectangular court-vard is 42 feet 9 inches long, and 
68 feet 6 inches in depth, and a foot-way goes quite through 
it, of 5 feet 6 inches in breadth ; the foot-way is laid with 
stone at 35. Qd. per yard, and the rest with pebbles at 3s. 
per yard : what will the whole come to 1 

Ans. 49Z. lis. Qkd* 



VAULTED AND ARCHED ROOFS. 



Arched roofs are either vaults, domes, saloons, or groins. 

Vaulted roofs are formed by arches springing from the 
opposite walls, and meeting in a line at the top. 

Domes are made by arches springing from a circular or 
polygonal base, and meeting in a point at the top. 

Saloons are formed by arches connecting the side walla 
to a flat roof, or ceiling, in the middle. 

Groins are formed by the intersection of vaults with 
each other. 

Domes and saloons rarely occur in the practice of meas- 
uring, but vaults and groins over the cellars of most houses. 

Vaulted roofs are generally one of the three following 
sorts : 

1. Circular roofs, or those whose arch is some part of 
the circumference of a circle. 

2. Elliptical roofs, or those whose arch is some part of 
the circumference of an ellipsis. 

3. Gothic roofs, or those which are formed by two cir- 
cular arcs that meet in a point directly over the middle of 
tlie breadth, or span of the arch. 

PROBLEM I. 

To find the solid content of a circular, elliptic, or gothle 
vaulted roof. 



YAVhTmi AN© ARCHEO BOOTS* 195 

RULE,* 

Multiply the area of one end by the length of the roo^ 
and the product will be the solidity required. 

EXA3IPLES. 

1. What is the solid content of a semi-circular vault 
whose span is 40 feet, and its length 120 feet? 
.7854 
WOO ^square of ^0, 

4712400 

7854 



2)1256.6400 

628.33^ area of the end. 
l%0 z^length. 



7 5S98A0== solidity required^ 

2. Required the solidity of an elliptic vault, whose span 
is 40 feet, height 12 feet, and length 80 ? 

Ans. 30159.36 /ee^. 

8, What is the solid content of a gotliic vault, whose 
gpan is 48, the chord of its arch 48, the distance of the 
arch from the middle of the chi^rd 18, and the length of the 
vault 60 ] Ans* 136224,71712.t 

PROBLEM IL 

Tofnd thecmieave, or convex surface, of circulary elliptic ^ 
or gothic vaulted roofs, 

* Tojind the solidity of the materials in either of the arches. 

Rule. From the solid content of the whole arch take the solid 
content of the void space, and the remainder will be the solidity of the 
arch. 

t The areas of these segments were calculated by rule 2, Prpfc, 
XIII. and tlie triangle by Pj-ob, VIIL 



196 VAULTED AKB ARCBffiD ROOFF. 

RULE* 

Multiply the length of the arch by the length of the vault, 
and the product will be the superficies required. 



What is the concave surface of a semi-circular vault, 
whose span is 40 feet and its length 120 1 

3.1416 
40 



2)125.6640 



62.S32z=lengfk of the arch. 
120 



7539.840=co«care surface required* 



PROBLEM m. 



To find the solid content of a dome ; its height y and the di' 
mensions of its base being known, 

RULE.t 

Multiply the area of the base by two-thirds of the height, 
and the product will be the solidity. 

EXAHFLES. 

1. What is the solid content of a spherical dome, the 
diameter of whose circular base is 60 feet? 



* The convex surface of a vault may be found by stretching a 
string- over it ; but for the concave surface tJiis method is not apphca- 
ble, and therefore its length must be found from proper dimensions. 

t Domes and saloons are of various fignres, but they are things that 
seldom occur in the practice of measuring. 



VArtTE© AI^D ARCHED ROOFS, 197 

.7854 
S60Q^ square of 00. 



4712400 
23562 



a827.4400= area of the base, 

20=1 of the height (30). 



56548. 8000 ==5o/i«?i^t/ required, 

2, In an hexagonal spherical dome, one side of the base 
is 20 feet; what is the solidity? Ans. 12000/ee^. 

PROBLEM IV. 

To find the superficial content of a spherical dome, 

RULE,* 

Multiply the area of the base by 2, and the product will 
be the superficial content required. 

EXAMPLES, 

1. What will the painting of an hexagonal spherical 
dome come to, at Is. per yard ; each side of the base being 
20 feet? 

2.598076=arca of a hexagon whose side is 1. 
4:00-z8quai'e of 20, 

1039.230400= area of the base, 
2 



9)207 8,4,60800= superficial content required^ 
2.0)230.940088 



11,5470044= llZ. 105. lid, the expense of painting, 

* The practical rule for elliptical domes is as follows : 
Rule. Add the height to half the diameter of the base, and this 
iBum multiplied by 1.5708 will give the superficial content nearly, 
R2 



198 VAULTED AXD ARCHED KOOFS. 

PROBLEM V. 

To find the solid content of a saloon* 

RULE.* 

1. Multiply the height of the arc, its projection, one- 
fourth of the perimeter of the ceiling, and 3.1416 continu- 
ally together, and call the product A. 

2. From a side or diameter of the room take a like side 
or diameter of the ceiling, and multiply the square of the 
remainder by the proper factor, (page 63,) and this product 
again by two-thirds of the height, and call the last product 
B. 

3. Multiply the area of the flat ceiling by the height of 
tlie arch, and this product added to the sum of A and B 
will give the content required. 



1. What is the solid content of a saloon with a circular 
quadrantal arch of 2 feet radius, springing over a rectangu- 
lar room of 20 feet long, and 16 feet wide ? 



* To^nd the superficial content of a saloon. 

1. Find the area of the flat part of the ceiling. 

2. Find the convex surface of a cylinder, or cylindroid, whose 
length is equal to onc-^ourth the perimeter of the ceiling, and its di- 
ameters to twice the height and twice tlie projection of tlie arch. 

3. Find the superficial nontcnt of a dome of the same figure as the 
n.rch, and whose base is either a square, or a figure similar to that of 
the ceiling; the side being equal to the difference of a side of the 
room and a side of the ceiling. 

4. Add these three articles together, and the sum will give the su- 
perficial content required. 

Note. — In a rectangular, circular, or polygonal room, the base or 
tlie dome will be a square, a circle, or a like polygon. 



VAULTED AND ARCHED ROOFS. 199 

Here the Jlai part of the ceiling ts 16 feet hy \2; and 

4)56 

14=5 of the perimeter, 
2=height. 

28 
2=prqjecti(m, 

56 
3.1416 
56 



188496 
157080 



175.9296=A. 



20= 
16= 


= side 
=side 


of 


the room, 
the ceiling. 


4 
4 








16 

1.000, i 


fee. 

=1 


=factor. 


16.000 
1^ 


of the height. 



16.000 
5.333 

21.333=B. 
16 
12 

192:^ area of the flat ceiling* 
2=height of the arch. 

384 



200 VAULTED AND ARCHED ROOFS. 

384 

175.9296 
21.3333 



581, 2Q29z= solid content required* 

2. A circular building of 40 feet diameter, and 25 feet 
high to the ceiling, is covered with a saloon, whose circu- 
lar quadrantal arch is 5 feet radius; required the capacity 
of the room in cubic feet. Ans. 30779.45948 feet. 



PROBLEM VI. 

To find the solid content of the vacuity formed by a groin 
arch, either circular or elliptical. 



RULE. 

Multiply the area of the base by the height, and the pro- 
duct again by .904, and it will give the solidity required. 

EXAMPLES. 

1. What is the solid content of the vacuity formed by a 
circular groin, one side of its square base being 12 feet? 

Here \2- X 6x .904=781.056=5oZi<?% required, 

2. What is the solid content of the vacuity formed by an 
elliptical groin, one side of its square base being 20 feet, 
and the height 6 feet ? Ans. 2169.6. 

PROBLEM Yll, 
^Tofind the concave superficies of a circular groin. 



VAULTED AND ARCHED ROOFS. 201 

RULE.« 

Multiply the area of the base by 1.1416, and the product 
will be the superficies required. 



* This rule may also be observed in elliptical groins, the error being 
too small to be regarded in practice. 

In measuring works where there are many groins in a range, the 
cylindrical pieces between the groins, and on their sides, must be 
computed separately. 

And to find the solidity of the brick or stone work, which forms the 
groin arches, observe the following 

RiTLK. Multiply the area of the base by the height, including the 
work over the top of the groin, and this product lessened by the solid 
content, found as before, will give the solidity required. 

Tlie general rule for measuring all arches^ is this: 

From the content of the whole, considered as solid, from the spring- 
ing of the arch to the outside of it, deduct the vacuity contained be- 
tween the said springing and the under side of it, and the remainder 
will be the content of the solid part. 

And because the upper sides of all arches, whether vaults or groins, 
are built up sohd, above tho haixncbea, to the Bcme height with the 
crown, it is evident that the area of the base will be the whole content 
above mentioned, taking for its thickness the height from the spring- 
ing to the top. And for the content of the vactdty to be deducted, take 
the area of its base, accounting its thickness to be two-thirds of the 
greatest inside height. But it may be noted that the area used in the 
vacuity, is not exactly the same with that used in the solid ; for the 
diameter of the former is twice the thickness of the arch less than 
that of the latter. 

And although I have mentioned the deduction of the vacuity as com- 
mon to both tie vault and the groin, it is reasonable to make it only in 
the former, on account of the waste of materials and trouble to the 
workman, in cutting and fitting them for the angles and intersections. 

Whoever wishes to see this subject more fully handled, may consult 
La Thiorie et la Pratique de la Geometric, par M. VAhM Deidier , 
a work in which several parts of Mensuration and Practical Geometry 
are skilfully handled, the examples being mostly wrought out in an 
easy femiliar manner, and illustrated with observations, and figures 
very neatly executed. 



202 VAriTEX) AND ARCHED ROOFS* 



EXAMPLES. 

1. What is the curve superficies of a circular groin arch, 
one side of its square being 12 feet? 

Here 122 x 1.1416 = 164. 3904 =«/^er/?cie* required. 

2. What is the concave superficies of a circular groin 
arch, one side of its square being 9 feet 1 Ans. 92.4696. 



CARPENTER'S RULE. 



This instrument is commonly called CogeshalPs sliding 
rule. It consists of two pieces, of a foot in length each, 
which are connected together by means of a folding joint. 

On one side of the rule, the whole length is divided into 
inches and half quarters, for the purpose of taking dimen- 
sions. And on this face there are also several plane scales, 
divided by diagonal lines into twelve parts, which are de- 
signed for planning such dimensions as are taken in feet 
and inches. 

On one part of the other face there is a slider, and four 
lines marked A, B, C, and D ; the two middle ones B and 
C being upon the slider. 

Three of these lines A, B, C, are double ones, because 
they proceed from 1 to 10 twice over: and the fourth line 
D is a single one, proceeding from 4 to 40, and is called 
the girth line. 

The use of the double lines A, and B, is for working 
proportions, and finding the areas of plane figures. And 
the use of the girth line D, and the other double line C, is 
for measuring solids. 

When 1 at the beginning of any line is counted 1, then 
the 1 in the middle will be 10, and the 10 at the end 100. 
And when 1 at the beginning is counted 10, then the 1 in 
the middle is 100, and the 10 at the end 1000, &c. and all 
the small divisions are altered in value accordingly. 



204 carpeinter's rule. 

Upon the other part of this face, there is a table of the 
value of a load of timber, at all prices, from 6d. to 2s. a 
foot. 

Some rules have likewise a line of inches, or a foot di- 
vided decimally into 10th parts; as well as tables of board 
measure, &;c. but these will be best understood from a sight 
of the instrument. 

THE USE OF THE SLIDING RULE. 

PROBLEM I. 
To find the product of two numbers, as 7 and 26. 

RULE. 

Set 1 upon A, to one of the numbers (26) upon B ; then 
against the other number (7) on A, will be found the pro- 
duct (182) upon B. 

Note. — If the third term runs beyond the end of the line, 
seek it on the other radius, or part of the line, and increase 
the product 10 times. 

PROBLEM n. 

7b divide one number by another, as 510 by 12. 

RULE. 

Set the divisor (12) on A, to 1 on B; then against the 
dividend (510) on A, is the quotient (42^) on B. 

Note. — l( the dividend runs beyond the end of the line, 
diminish it 10 or 100 times to make it fall on A, and in- 
crease the quotient accordingly. 

PROBLEM in. 

To square any number, as 27. 

RULE. 

Set 1 upon D to 1 upon C ; then against the number 
(27) upon D, will be found the square (729) upon C- 



carpenter's rule. 205 

If you would square 270, reckon the 1 on D to be 100 ; 
and then the 1 on C will be 1000, and the product 72900. 



PROBLEM IV. 

To extract the square root of any number, as 4268. 

RULE. 

Set 1 upon C, to 1 upon D ; then against (4268) the 
number on C, is (65.3) the root on D. 

To value this right, you must suppose the 1 on C to be 
some of these squares, 1, 100, 1000, &c. which is the 
nearest to the given number, and then the root corre- 
sponding will be the value of the 1 upon D. 

PROBLEM V. 

To find a mean proportional between any two nvmberSy a>s 
27 and 450. 

RULE. 

Set one of the numbers (27) on C, to the same on D , 
then against the other number (450) on C, will be the 
mean (110.2) on D. 

Note. — If one of the numbers overruns the line, take the 
lOOtb part of it, and augment the answer 10 times- 

PROBLEM VI. 

Three numbers being given, to find a fourth praporiianal ; 
suppose 12, 28, and 57. 
S 



206 CARPENTERS RULE. 

Set the first number (12) upon A, to the second (28) 
upon B ; then against the third number (57) on A, is the 
fourth (133) on B. 

Note* — If one of the middle numbers runs off the line, 
take the tenth part of it only, and augment the answer 10 
times. 

The finding a third proportional is exactly the same, the 
second number being twice repeated. 

Thus, suppose a third proportional was required to 21 
and 32. 

Set the first 21 on B, to the second 32 on A; then 
against the second 32 on B, is 48.8 on A, which is the third 
proportional required. 



* The use of the rule in board and timber measure will be shown 
in what follows. 

If the breadth of a board be given ; tojind how much in length will 
make a square foot. 

Rule. If the board be narrow, it will be found in tlie table of 
board measure on the rule ; but, if not, shut the rule, and seek the 
breadth in the line of board measure, running along tlie rule, frora 
that table ; then over against it, on the opposite side, is the length in 
inches required. 

The side of the square of a piece of timber being given ; tojind how 
much in length will make a foot solid. 

Rule. If the timber be small, it will be found in the table of tim- 
ber measure on the rule; but, if not, look lor the side of the square, 
in the line of timber measore, running along the rule, from that table, 
and against it in the line of inches is the length required. 



OP 

TIMBER MEASURE, 



PROBLEM I. 
Toj^nd the areu, or superficial content of a hoard or planh^ 

RULE, 

Multiply the length by the breadth, and the product 
will be the content required, 

iVbie.— When the board is tapering, add the breadths of 
the two ends together, and take half the sum for the mean 
breadth. 

BY THE SLIDING RULE. 

Set 12 on B to the breadth in inches on A, then against 
the length i-n feet on B is the content on A, in feet and 
fractional parts as required. 

EXAMPLES. 

1. What is the value of a plank, whose lengtb is 8 feet 
6 inches, and breadth 1 foot 8 inches throughout, at 2|<?* 
per foot ? 

feet in, 
8 6 

1 3 

8 6 

2 16 



10 7 6 the eontent* 



208 



TIMDEE 


MEASURE, 




10 


7 6 


2d, is i 


1 


8 


*isi 




5 


m. 






6isi 




U 


lis| 




i 



2«, 6|c?« the Answer* 

BY THE SLIDING RULE. 

As 12 on B : 15 on A : : 8^ on B ; lOl on A. 

2. What is the content of a board, whose length is 5 
feet 7 incheSj and breadth 1 foot 10 inches? 

Ans. 10/e. 2 in. 10 pa. 

3. At l-|<f. per foot, what is the value of a plank whose 
length is 12 feet 6 inches, and breadth 11 inches througU- 
out? Ans. Is, bd. 

4. Find the value of 5 oaken planks at M. per foot, 
each being 17^ feet long, and their particular breadths as 
follows : viz. two of IS^ inches in the middle, one of 14^ 
inches in the middle, and the two remaining ones, each 18 
inches at the broader end, and 11^ at the narrower? 

Ans. IZ. 55, 9i<?. 

PROBLEM II. . 
Tojind the solidity of squared or four-sided timher<. 

RULE.* 

Multiply the mean breadth by the mean thickness, and 
this product again by the length, and it will give the so- 
lidity required. 

* Note 1. If the stick be equally broad and thick throughout, the 
breadth and thickness, anywhere taken, will be the mean breadth 
and thickness. 



TmBER JtEAStTRB. 509 



BY THE SLIDING RULE. 



As the length in feet on C ; 22 on D ; : quarter girth in 
inches on D : solidity on C, 



1. The length of a piece of timber is 20 1' feet, the 
breadth at the greater end is i foot 9 inches, and the thick- 
ness 1 foot 3 inches ; and at the less end the breadth is 1 
foot 6 inches, and the thickness 1 foot : what is the solidity? 

l. "15 7=^ greater breadths 
1.5 =iless hreadthc 

^)3.25 

l.Q25='niean breadth* 



2. If the tree tapers reg-ularly from one end to the other, the 
breadth and thickness, taken iji tlie middle, will be the mean lu-eadth 
and thickness, 

3. If the stick does not taper re^larly, but is tliicker in SQin§ 
places than in others, let several different dimensions be taken, and 
their sura divided by the nuinber of them will give the mean dimen.- 
sions. 

This method of finding- the mean dimensions is mostly used in 
practice, but, in many cases, it is exceedingly erronecus. 

The quarter girth, likewise, which is mentioned in the proporticai 
by the sliding rule, is subject to error. It is not the fourth part of the 
circumference, but tlie square root of the product arising from inultL 
plying the mean breadth by the mean thickness. 

In order to show the fallacy of taking one^fotirth of the girth for 
tlie side of a mean square, tai:e the following example ; 

Suppose a piece of timber to be 24 feet long, and a foot square 
throughout, and let it be slit into two equal parts, from end to end. 

Then the sum o^'the solidities of the two parts, by the quarter girth 
method, will be §7 feet, but the true solidity is 24 feet ; and if the tW9 
pieces were very unequal, the difference would be still greater, 

S3 



210 TI3IBER MEASTHB. 

1.2^^= greater thickness. 
1.00 = l€ss thickness, 

2)2.25 

1.1 25= mean thickness. 
Now 1.625X 1.125X 20.5=37.4765625=co«^e?i< required. 

By Cross Multiplication, 

fe. in. pa, 
1 7 6 
116 



17 6 

1 7 
9 


6 
9 




1 9 11 
20 6 


3 




36 6 9 
10 11 




7 


6 



37 5 8 7 6=c<mt€nt, 



BY THE SLIDING RULE. 



As 1 upon B : 19y«2 upon A :: 13/^ upon B : 263yVT 
upon A, the mean square, 

^5 16 upon C : 4 z/pon D : : 1.8 upon C : 16.2 upon D, 
the side of the mean .square. 

As 20 1 upon C : 12 upon D : : 16.2 upon D : 37y'j upon 
C, the ansxner, 

2. The length of a piece of timber is 24.5 feet, and its 
ends are equal squares, whose sides are each 1.04 feet: 
what is the solidity? Ans. 20 feet 6 inches, 

3. The length of a piece of timber is 20.38 feet, and 
the ends are uiiequal squares, the side of the greater being 



TIMBER MEASURE. 211 

19i inches, and that of the less 9| inches : what is the so- 
lidity? Ans. 29.7 5Q feet. 
4. The length of a piece of timber is 27.36 feet; at the 
greater end the breadth is 1.78 feet, and the thickness 1.23 
feet; and at the less end the breadth is 1.04 feet, and the 
thickness .91 feet : what is the solidity ? 

Ans. 41.278/ce^ 

PROBLEM HI. 
To find the solidity of round or unsquared timber* 

RULE 1.* 

Multiply the square of the quarter girth (or one-fourth 
of the circumference) by the length, and the product will 
be the content, according to the common practice. 



* Let c= girth or circumference, and Z= length of the tree. 

c c cH 

Then X^X^^~ 16" content of the tree according 

to the rule. 

^^^ 4x3 1416 '^ ^ ~ 19 5664 ~ ^^"^ content, accord- 
ing to the rule for finding the content of a cylinder. 
cH . cH 

But TO 5664 (differs from y^ by nearly one-fourth part of 

the whole, and therefore the rule is exceedingly erroneous. 

When the tree is tapering, the mean girth is found in the same 
maimer as in board measure. Or if the tree be very irregular, the 
best way is to di\'ide it into a certain number of lengths, and find the 
content of each part separately. 

When trees have their bark on, an allowance is generally made, by 
deducting so much from the girth as is judged sufficient to reduce it 
to such a circumference as it would have without its bark. In oak 
this allowance is about jq or Jj part of the girth; but for elm, beach, 
ash, &c. whose bark is not so thick, the deduction ought to be less. 



212 



TIMBER MEASUKB, 



BY THE SLIDING HCLE. 



Ai the length upon C : 12 upon D :: \ girth upon D : 
content upon C. 

EXAMPLES. 

1. A piece of timber is 9| feet long, and the quarter 
girth is 39 inches ; what is the solidity ? 

By Decimals, 
Here S9in.=S.25ft., and 9f /^=9.75/^ 
Hence 3.25{2x 9.75 = 10.5625 X 9.75=102.984375 cubic 
feet= solidity required. 

By Cross Multiplication, 



ft. in, 
3 3= 
3 3 

9 9 
9 


= 39 inches. 
9 


10 6 
9 9 


9 
= 9| feet. 


95 
7 11 


9 
9 


102 ii 


9 9=solid 



BY THE SLIDITVCJ RULE. 

Ai 9| upon C : 12 upon D : : 39 ujyon D : 103 upon C, 
the content. 

2. Tlie length of a tree is 25 feet, and the girth through- 
out 2^ feet; what is its soliditv ? Ans. 9 feet 9 inches, 

3. The length of a tree is 14| feet, and its girth in the 
middle 3.15 feet; required the solidity? 

Ans. 9 feet nearly. 



TmBER arEASlTKE, 213 

4. The girths of a tree in 4 different places are as foU 
lows : in the first place 5 feet 9 inches, in the second, 4 
feet 6 inches, in the third 4 feet 9 inches, and in the fourth 
3 feet 9 inches ; and the length of the whole tree is 15 feet; 
what is the solidity 1 Ans. 20 feet 7 ijiches* 

5. An oak tree is 45 feet 7 inches long, and its quarter 
girth 3 feet 8 inches ; what is the solid content, allowing 
j\ for the bark ? Ans* 5l5i feet ^ nearly* 

mjLE 11,* 

Multiply the square of one-fifth of the girth hy,twic^^^|^ 
the truths 



* Let c 5s= circumference, and Z== length, as before* 

c c 2cH cH 

Then -^-x y X 2? = 205— X25=^Q^*®^* ^^ ^^^ ^^^^ ^^* 
cording to the rule. 

And the true content is = 12 5664 ' ^^ ^^® before shown. 
cH cH 1 

But Y2^ differs from y2.bm4. ^^ ^"^^ ^^^^^ 190 P^^^ ^^ 
the whole, and is therefore sufficiently near the truth for any practical 
purpose. 

This rule is full as easy in practice as the false one, and therefor© 
ought to be generally used, since the ease of the other method is the 
xmly argument which is alleged for em^ploying it. 

The following rule was given me by Mr. Burrow^ and is a still 
nearer approximation. 

Rule. Multiply the square of the circumforence by the length, 
and take y\ of the product : from this last number subtract ■g' of it- 
self, and the remainder wiU be the answer. 

c^l 1 8 1 

^^^ 12:5664=88 ^^ ^^ ^^^ "^'^^^^Z. = (88-~88> ^ ^^ 

c"! 1 cH 
^r-r- — -^ of jT, which is the same as the rule, and differs 

from the truth by only 1 foot in 2300, 



214 TIMBER MEASURE, 



BY THE SLIDING RULE. 



As twice the length upon C : 12 upon D : : one-fifth of 
the girth upon D : content upon C. 



EXAMPLES. 

1. A piece of timber is 9f feet long, and | of the girth 
is 2.6 feet ; what is the solidity ? 

iJy jjecimats, 

2.6 
2.6 

156 
52 

6.76 
9.75 

3380 
4732 
6084 

65.9100 
2 

Here'2Sfx 9.75 X 2 = 121.8200= content. 



BY THE SLIDING RULE. 

As 19.15 upon C : 12 upon D : : 31i in. upon D : 132 
the content upon C. 

2. If the length of a tree be 24 feet, and the girth 
throughout 8 feet ; what is the content ? 

Ans. 12S feet, nearly* 



TIMBER MEASURE. 215 

3. If a tree girth 14 feet at the thicker end, and 2 feet at 
the smaller end ; required the solidity when the length is 
24 feet ? Ans. 1 2 3 feet, nearly, 

4. A tree girths in five different places as follows : in the 
first place 9.43 feet, in the second 7.92 feet, in the third 
6.15 feet, in the fourth 4.74 feet, and in the fifth 3.16 feet; 
and the whole length is 17^ feet; what is the solidity? 

Ans. 54.4249/ec<. 



SPECIFIC GRAVITY. 



The specific gravities of bodies are their relative weights 
contained under the same given magnitude, as a cubic foot, 
a cubic inch, &c. 

The specific gravities of several sorts of bodies are ex- 
pressed by the numbers annexed to their names in the fol- 
lowing table : 



A table of the specific gravities of bodies* 

Brick - . . 
Light earth - - 
Solid gunpowder 
Sand .... 
Pitch .... 
Dry box-wood - 
Sea water - - 
Common water - 
Dry oak ... 
Gunpowder, shaken 
Dry ash - - ■ 
Dry maple - • 
Dry elm - - • 
Dry fir - - . 
Cork .... 
Air ... . 



Fine gold - - ■ 


19640 


Standard gold - • 


18888 


Quicksilver 


14000 


Lead . . . . 


11325 


Fine silver - • 


11091 


Standard silver • 


10535 


Copper . - - 


. 9000 


Gun metal 


. 8784 


Cast brass - - 


. 8000 


Steel - - . 


. 7850 


Iron - - - 


. 7645 


Cast iron - - 


. 7425 


Tin ... - 


. 7320 


Marble - - - 


. 2700 


Common stone 


2520 


Loam - - . . 


2160 



2000 

1984 

1745 

1520 

1150 

1030 

1030 

1000 

925 

922 

800 

755 

600 

550 

240 

li 



Note. — As a cubic foot of water weighs just 1000 
ounces Avoirdupois, the numbers in this table express not 
only the specific gravities of the several bodies, but also 



SPECIFIC GKAVITY. 217 

the weight of a cubic foot of each, in Avoirdupois ounces; 
and hence, by proportion, the weight of any other quantity, 
or the quantity of any other weight, may be readily known. 



PROBLEM I. 

To find the magnitude of a body, from its weight being 
given, 

RULE. 

As the tabular specific gravity of the body, is to its 
weight in Avoirdupois ounces. 

So is one cubic foot, or 1728 cubic inches, to its content 
in feet, or inches, respectively. 



EXAMPLES. 

1. Required the content of an irregular block of com- 
mon stone, which weighs 1 cwt. or 112 lbs. 

Here 112 lbs.=1792 oz. 

oz. oz. c. in. c. in. 
Hence 2520 : 1792 : : 1728 : 1228a. Ans. 

2. How many cubic inches of gunpowder are there in 
one pound weight? Ans. 30 nearly. 

3. How many cubic feet are there in a ton weight of 
dry oak? Ans. 38if|. 



PROBLEP>I n. 

To find the weijght of a body from its Tnagnitude being 
given. 
T 



218 SPECIFIC GBAVITY. 



RULE. 



As one cubic foot, or 1728 cubic inches, is to the con- 
tent of the body. 

So is its tabular specific gravity, to the weight of the 
body. 

EXAMPLES. 

1. Required the weight of a block of marble, whose 
length is 63 feet, and its breadth and thickness each 12 
feet ; these being the dimensions of one of the stones in 
the walls of Balbec. 

Here 12' x 63=9072 c. ft.=content of the body. 

Hence 1 : 9072 : : 2700 : 244944 oz.=638y\ tms, which 
is equal to the burthen of an East India ship, 

2. What is the weight of a pint of gunpowder, ale 
measure ? Ans. 19 oz. nearly. 

3. What is the weight of a block of dry oak, which 
measures 10 feet in length, 3 feet in breadth, and 2^ feet 
deep? Ans. 4335if lbs, 

PROBLEM III, 

Tojind the specific gravity of a body, 
RULE. 

Case 1. When the body is heavier than water, weigh it 
both in water and out of water, and the difference will be 
the weight lost in the water. 

Then, as the weight lost in the water, is to the whole 
weight, 

So is the specific gravity of water, to the specific gravity 
of the body. 



SPECIFIC GRAVITY^ 219 



A piece of stone weighed in air 10 pounds, but in water 
only 6| pounds. Required its specific gravity. 



10 
61 




H 

13 : 


10 : : 1000 : 
40 : : 1000 : 
40 




13)40000(3077 
39 




100 
91 



90 

Case 2, When the body is lighter than water, so that it 
will not quite sink, affix to it another body heavier than 
water, so that the mass compounded of the two may sink 
together. 

Weigh the heavier body and the compound mass sepa- 
rately both in water and out of it, and find how much each 
loses in water, by subtracting its weight in water from its 
weight in air. 

Then as the difference in these remainders is to the 
Weight of the light body in air. 

So is the ^ecific gravity of water to the specific gravity 
of the body.. 



Suppose a piece of elm weighs in air 15 pounds, and 
that a piece of copper which weighs 18 pounds in air, and 
16 pounds in water, is affixed to it, and that the compound 
weighs 6 pounds in water | required the specific gravity ,Qf 
the elnu 



220 SPECIFIC GRAVITY. 

18 in air 33 
16 in water 6 

loss 2 27 loss. 

2 

25 
25 : 15 : : 1000 : 600 Abs. 

PROBLEM IV. 

To find the quantities of tico ingredients in a given 
compound, 

RULE. 

Take the difference of every pair of the three specific 
gravities, viz. of the compound and each ingredient, and 
multiply the difference of every two by the third. 

Then as the greatest product is to the whole weight of 
the compound, so is each of the other products to the 
weight of the two ingredients. 



A composition of 112 pounds being made of tin and 
copper, whose specific gravity is found to be 8784 ; re- 
quired the quantity of each ingredient : the specific gravity 
of tin being 7320, and of copper 9000. 

9000 9000 8784 

7320 8784 7320 



1680 

8784 


216 
7320 


1464 diff. 
9000 


702720 
52704 

8784 


4320 
648 
1512 


13176000 


14757120 


1581120 





SPECIFIC GRAVITY. 221 

14757120 : 112 : : 13176000 
112 



26352000 
13176 
13176 



14757120)1475712000(100 

Ans. 100Z&. of copper > . .i,. comvosition 
and I2lb. of tin ] '"* ^^^ composition. 



T8 



MISCELLANEOUS QUESTIONS. 



1. What difference is there between a floor 48 feet 
long, and 30 feet broad, and two others each of half the 
dimensions ? Ans. 120 feet. 

2. From a mahoo^any plank 26 inches broad, a yard and 
a half is to be sawed off; what distance from the end 
must the line be struck? Ans. 6.23 feet. 

3. A joist is 8-2- inches deep, and 3| broad ; what will be 
the dimensions of a scantling just as big again as the joist, 
that is 4| inches broad? Ans. 12.52 inches deep. 

4. A roof is 24 feet 8 inches by 14 feet 6 inches, and is 
to be covered with lead at 8ibs. to the foot ; what will it 
come to at I85. per cwt. ? Ans. 22Z. 195. lO^d. 

5. What is the side of that equilateral triangle, whose 
area cost as much paving at 8d. per foot, as the palisading 
the three sides did at a guinea per yard 1 

Ans. 72.7^6 feet. 

6. The two sides of an obtuse-angled triangle are 20 
and 40 poles; what must the length of the third side be 
that the triangle may contain just an acre? 

Ans. 58.876, or 23.099. 

7. If two sides of an obtuse-angled triangle, whose area 
is 60 v/ 3, be 12 and 20; what is the third side ? 

Ans. 28. 

8. If an arra of 24 be cut off from a triangle, whose 
three sides are 13, 14, and 15, by a line paralh 1 to the 
longest side; what arc the lengths of the s!(hs including 
that area? Ans. y 7 14, 2 V 14, and y ^ iZ 

9. The distance of the centres of two circlrs, whose 
diameters are each 50, is equal to 30 ; whrt is the area of 
the space inclosed by their circumference ? 

Ans. 559.115. 



MISCELLANEOUS QUESTIONS. 



223 



10. There is a segment of a circle the chord of which 
is 60 feet, and versed sine 10 feet ; what will be the versed 
sine of that segment of the same circle whose chord is 90 
feet? Ans. 28.2055. 

11. The area of an equilateral triangle, whose base falls 
on the diameter, and its vertex in the middle of the arc of 
a semicircle, is equal to 100; what is the diameter of the 
semicircle 1 Ans. 26.32148. 

12. The four sides of a field, whose diagonals are equal 
to each other, are 25, 35, 31, and 19 poles, respectively; 
what is the area?* Ans. 4 ac. 1 ro. 38 poles, 

* Construction. In this question the sums of the squares of the 
opposite sides of the trapezium being equal, the figure may be con- 
structed as follows : 

E 




Draw AB and AE at right angles, and each equal to one of the 
given sides (33) ; join BE, and from the points E and B, with radii 
equal to the two remaining opposite sides (25 and 31) respectively, 
describe arcs intersecting in C on the farther side of BE; join AC, 
and draw BF at right angles to it. With the centre C, and radius 
equal to the remaining side (19) describe an arc cutting BF produced 
in D. Join AD and CD, then will ABCD be the figure required. 

Demonstration.'— By the question ab^ + cd^ = bc^ + ce^ 
and since bd and ac cross each other at right angles, 
(47.1) AB" -{- CD' = Bc^ + AD^ ; whercforc ad^=ec , or ad 

= EC. 



224 MISCELLANEOUS QUESTIONS. 

13. A cable which is 3 feet long, and 9 inches in com- 
pass, weighs 22 pounds : what will a fathom of that cable 
weigh whose diameter is 9 inches 1 Ans. 434.25 lbs. 

14. A circular fish-pond is to be dug in a garden that 
shall take up just half an acre : what must the length of the 
chord be that strikes the circle ? Ans. 27.75 yards. 

15. A carpenter is to put an oaken curb to a round 
well, at Sd, per foot square ; the breadth of the curb is to 
be 7^ inches, and the diameter within 3^ feet : what will 
be the expense? Ans. 5s. 2\d. 



Hence, in the two triangles ABD and EAC, we have the two sides 
BA, AD equal the two AE, EC, each to each, and the angles ABD 
and EAC equal (each being the complement of BAF) and EC and 
AD, similarly situated; wherefore BD=AC. Q. E. D. 

Calculation. — On be let fall the perpendiculars cg and 
AH ; now BE^ =AB^+AE^=35^ X 2 ; be = -v/ 35^ X 2 =35 
v/2=49.4975; and AH=:BH=iBE=24.7487 ; again (13.2) 
bc' + be^— CE^ 31^ + 2 X 35^—25' 2786 

BG = 



2be 2 X 49.4975 98.995 

28.1428; gh = bg — bh = 28.1428 — 24.7487 =3.3941 J 



cg=v/bc' — bg'=x/ 312—28.1428^= v/ 168.98280816 = 
12.9993. By sim. triangles ah + cg(37.748) : gh(3.3941) 
: : AH (24.7487) : hi = 2.2253; ai = V ah^ + ni^ = 



v/24.74872+ 2.2253^ = V 612.49815169 + 4.95196009 = 
v/ 617.45011178=24.8485. Again, by sim. triangles, m 
(2.2253) : hg(3.3941) : : ai(24.8485) : ac=37.8997 = bd ; 
now, by Problem V. Superficies, the area of the trape- 
acXbf + fd acXbd AC" 37.8997^ 

Zium AECD =: ^ = ^ = -^ = ^ ^^ 

718.1936 po. = 4 ac. 1 ro. 38 po. Ans. 

Note. — ^This method is applicable to all questions of the kind 
wherein the diagonals cross each other at right angles ; that is, when 
the sums of the squares of the opposite sides are equal, but a general 
solution to the question, without tliis, w^ouid involve an equation of the 
higher order. 



MISCEIlLANEOUS aUESTIONS. 



225 



16. Suppose the expense of paving a semicircular plot, 
at 2s. 4;d, per foot, amounted to lOZ. what is the diameter 
of it? Ans. 14.7739. 

17. Seven men bought a grinding stone of 60 inches in 
diameter, each paying | part of the expense ; what part of 
the diameter must each grind down for his share 1 

C The 1st, 4.4508 ; 2d, 4.8400 ; 3^^, 5.3535 ; 
Ans. ^ Ath, 6.0765; 5th, 7.2079; 6ih, 9.3935; 
f and the 1th, 22.6778. 



* This problem may be thus constructed : 

On the radius AC describe a semicircle ; also divide AC into as 
many equal parts CD, DE, EF, &c. as there are shares, and erect the 
perpendicular DL, EM, FN, &c. meeting the semicircle described on 
AC in L, M, N, O, P, Q. Then with the centre C and radii CL, CM, 
CN, &c. describe circles, and the thing is done. 




For, by the nature of the circle, the square of the chords or radii 
CL, CM, CN, &c. are as the cosines CD, CE, CF, &c. 



226 



MISCELLANEOUS QUESTIOKS. 



18. A gentleman has a garden 100 feet long, and 80 feet 
broad, and a gravel walk is to be made of an equal width 
half round it ; what must the width of the walk be so as to 
take up just half the ground ?* 

Ans. 25.9G88/6C^ 

19. In the midst of a meadow well stored with grass, 
I took just an acre to tether my ass ; 

How long must the cord be, that feeding all round, 
He may'nt graze less or more than an acre of ground ? 

Ans. 39.25073/ee^ 

20. A malster has a kiln that is 16 feet 6 inches square ; 
now he wants to pull it down, and build a new one that 
will dry three times as much at a time as the old one did ; 
what must be the length of its side ? 

Ans. 2Qfeet 7 inches. 

21. If a round cistern be 26.3 inches in diameter, and 
52.5 inches deep ; how many inches in diameter must a 
cistern be to hold twice the quantity, the depth being the 
same? Ans. 37.19 inches. 



* This problem may be constructed thus : Let ABCD represent the 
garden, make CE=CB, and with the centre D and radius DE de- 
scribe the semicircle GEF. Make BI= 4 BG, BL= ^ BF, and com- 
plete the rectangle IBLH, and the thing is done, 



^ 


\ 


IL 





A I B 

BG BF BD -f (CD— CB) 

For the area of bihl=bi.bl=-2— X -g" ~ o 

BD (cD— Cb) BD* (CD Cb)'^ Bd' (CD^ 2CD.CB -f CB^) 

2 "" 4 ^ ' 4 



BD^ 80^^+200.08 CD.CB 

= ^^~~2 — ~^ '^^ ^^^ of the garden. 

Q. E. D, 



MISCELLANEOTTS Q,T7ESTI01SS. 227 

23. A may -pole, whose top was broken off by a blast of 
wind, struck the ground at 15 feet distance from the bot- 
tom of the pole : what was the height of the whole may- 
pole, supposing the length of the broken piece to be 39 
feet ] Ans. lb feet.. 

24. What will the diameter of a globe be, when the 
solidity and superficial content thereof are represented by 
the same number 1 Ans. 6. 

25. How many three inch cubes can be cut out of a 12 
inch cube 1 Ans. 64. 

26. A farmer borrowed part of a hay-rick of his neigh- 
bor, which measured 6 feet every way, and paid him back 
again by two equal cubical pieces, each of whose sides was 
three feet : Query, whether the lender was fully paid ? 

Ans. He was paid ^ part only, 

27. What will the painting of a conical church spire 
come to at 8d. per yard ; supposing the circumference of 
the base to be 64 feet, and the altitude 118 feet? 

Ans. 14Z. 05. SU- 

28. What will a marble frustrum of a cone come to at 
125. per solid foot ; the diameter of the greater end being 
4 feet, that of the less end 1^ feet, and the length of the 
slant side 8 feet? Ans. 30Z. I5. ll^d. 

29. The diameter of a legal Winchester bushel is 18^ 
inches, and its depth 8 inches : what must the diameter of 
that bushel be whose depth is 7^ inches ? 

Ans, 19.1067. 

30. Suppose the ball at the top of St. Paul's Church is 6 
feet in diameter : what did the gilding of it come to at 
3^d. per square inch? Ans. 2377. IO5. Id. 

31. A person wants a cylindric vessel of 3 feet deep, 
that shall hold twice as much as a vessel of 28 inches deep, 
and 46 inches in diameter : what must be the diameter of 
the vessel required? Ans. 57.37 incites. 



228 MISCELLANEOUS QUESTIONS. 

32. Two porters agreed to drink off a quart of strong 
beer between .them, at two pull:., or a draugJit each; now 
the first having given it a black eye, as it is called, or 
drank till the surface of the liquor touched the opposite 
edge of the bottom, gave the remaining part of it to the 
other ; what was the difference of their shares, supposing 
the pot was the frustrum of a cone, the depth being 5.7 
inches, the diameter at the top 3.7 inches, and that of the 
bottom 4.23 inches ? Ans. 7.05 cubic inches. 

33. The monument erected in Babylon by Queen Semi- 
ramis at her husband Ninus's tomb, is said to have been 
one solid block of marble in the form of a pyramid • the 
base was a square whose side was 20 feet; and the height 
of the monument was 150 feet; now suppose this monu- 
ment was sunk in the Euphrates, what weiirht would be 
sufhcient to raise the apex of it to the surface of the 
water, and what weight would raise the whole of it above 
the water? 

Ans. To raise it to the surface wonU require 94.«3i tons to 
raise it out would require 1506|f tons, (which is 'the 
same as the weight of the monument.) 

34. If the pyramid described in the last example were 
divided into three equal parts by planes parallel to its base, 
what would be the length of each part, beginning at the 

Ans. 104.0042, 27.0329 and 18.9629 respectively. 

35. How high above the surface of the earth must a 
person be raised to see ^ of its surface ? 

Ans. To the height of the earth's diameter. 

36. A cubical foot of brass is to be drawn into a wire of 
4V of an inch in diameter; what will be the length of the 
wire, allowing no loss in the metal? 

Ans. 97784.5684 yards, or near 56 Tniles. 

37. A gentleman has a bowling green, 300 feet long, 
and 200 feet broad, which he would raise one foot higher 
by means of the earth to be dug out of a ditch that goes 
round It; to what depth must the ditch be dug, supposing 
Its breadth to be everywhere 8 feet ? 

Ajis. 1^1 feet. 



MISCELLANEOUS QUESTIONS. 229 

38. Of what diameter must the bore of a cannon be, 
which is cast for a ball of 24lbs. weight, so that the diame- 
ter of the bore may be j\ of an inch more than that of the 
ball ? Ans. 5.647 inches. 

39. One end of a certain pile of wood is perpendicular 
to the horizon, the other is in the form of an inclined 
plane : the length of the pile at the bottom is 64 feet, length 
at the top 50 feet, height 12 feet, length of the wood 5 
feet ; required the number of cords it contains ? 

Ans. 26||. 

40. The ellipse in Grosvenor-square measures 840 links 
across the longest way, and 612 the shortest, within the 
walls ; now the walls being 14 inches thick, it is required 
to find what ground they inclose, and what they stand upon. 

Ans. They inclose Aac. Oro. 6po. and stand on 1760^ 
square feet. 

41. If a heavy sphere whose diameter is 4 inches, be 
put into a conical glass, full of water, whose diameter is 5, 
and altitude 6 inches : it is required to know how much 
water will run over 1 

Ans. l^ of a -pint nearly, or 26.272 inches. 

42. Suppose it be found by measurement, that a man-of- 
war, with its ordnance, rigging and appointments, draws so 
much water as to displace 50,000 cubic feet of water; re- 
quired the weight of the vessel 1 

Ans. 1395/^ tons. 

43. One ev'ning I chanc'd with a tinker to sit, 
Whose tongue ran a great deal too fast for his wit : 
He talked of his art with abundance of mettle ; 
So I ask'd him to make me a flat-bottom'd kettle. 
Let the top and the bottom diameters be, 

In just such proportion as five is to three : 
Twelve inches the depth I proposed, and no more ; 
And to hold in ale gallons seven less than a score. 
He promised to do it, and straight to work went ; 
But when he had done it he found it too scant. 
U 



230 MISCELLA^^EO^S QUESTIONS. 

He alter'd it then, but too big he had made it; 
For though it held right, the diameters fail'd it ; 
Thus making it often too big and too little. 
The tinker at last had quite spoiled his kettle ; 
But declares he will bring his said promise to pass, 
Or else that he'll spoil every ounce of his brass. 
Now to keep him from ruin, I pray find him out 
The diameter's length, for he'll ne'er do 't, I doubt. 

Ans. The bottom diameter is 14.4401, and the top diame- 
ter 24.4002. 

44. If the above-mentioned frustrum of the cone were 
to hold as much again, what would be the length of the part 
added to the greater end ? 

Ans. 6.384 in. nearly. 



A 

TABLE 



OF THE 



AREAS OF THE SEGMENTS OF A CIRCLE, 

Whose diameter is Unity, and supposed to be divided into 1000 
equal Parts. 



1 § 


Seg. Area. 


li 


Seg. Area. 


! li 


Seg. Area. 


^m 




>m 




>j| 




.001 


.000042 


.024 


.004921 


! .047 


.013392 


.002' 


.000119 


.025 


.005230 


i .048 


.013818 


.003 


.000219 


1 .026 


.005546 


i .049 


.014247 


.004 


.000337 


.027 


.005867 


1 .050 


.014681 


.005 


,000470 


.028 


,006194 


.051 


.015119 


.006 


.000618 


.029 


.006527 


.052 


.015561 


.007 


.000779 


.030 


.006865 


.053 


.016007 


.008 


.000951 


.031 


.007209 


,054 


.016457 


.009 


.001135 


.032 


.007558 


.055 


.016911 


.010 


.001329 


.033 


.007913 


.056 


.017369 


.011 


.001533 


.034 


.008273 


,057 


.017831 


.012 


.001746 


.035 


.008638 


.058 


.018296 


.013 


.001968 


1 .036 


.009008 


.059 


,018766 


.014 


.002199 


.037 


.009383 


.060 


.019239 


.015 


.002438 


1 .038 


.009763 


.061 


.019716 


.016 


.002685 


.039 


.010148 


.062 


.020196 


.017 


.002940 


.040 


.010537 


1 ,063 


.020680 


.018 


.003202 


.041 


.010931 


1 .064 


.021168 


.019 


.003471 


.042 


.011330 


.085 


.021659 


.020 


.003748 


.043 


.011734 


,066 


,022154 


.021 


.004031 


.044 


.012142 


.067 


,022652 


.022 


.004322 


.045 


.012554 


.068 


.023154 


.023 


.004618 


, .046 


.012971 


.069 


.023659 



( 232 ) 



The Areas of the Segments of a Circle. 



>'U2 


Seg. Area. 




Seg. Area. 


It 


Seg. Area. 


.070 


.024168 


.102 


.042080 


.134 


.062707 


.071 


.024680 


.103 


.042687 


.135 


.063389 


.072 


.025195 


.104 


.043296 


.136 


.064074 


.073 


.025714 


.105 


.043908 


.137 


.064760 


.074 


.026236 


.106 


.044522 


.138 


.065449 


.075 


.026761 


.107 


.045139 


.139 


.066140 


.076 


.027289 


.108 


.045759 


\ .140 


.066833 


.077 


.027821 


.109 


.046381 


.141 


.067528 


.078 


.028356 


.110 


.047005 


1.142 


.068225 


.079 


.028894 


.111 


.047632 


.143 


.068924 


.060 


.029435 


.112 


.048262 


.144 


.069625 


.081 


.029979 


.113 


.048894 


.145 


.070328 


.032 


.030526 


.114 


.049528 


.146 


.071033 


.083 


.031078 


.115 


.050165 


.147 


.071741 


.034 


.031629 


.116 


.050804 


.148 


.072450 


.085 


.032186 


.117 


.051446 


.149 


.073161 


.0S6 


.032745 


.118 


.052090 


.150 


.073874 


.087 


.033307 


.119 


.052736 


.151 


.074589 


.088 


.033872 


.120 


.053385 


• .152 


.075306 


.089 


.034441 


.121 


.054036 


.153 


.076026 


.090 


.035011 


.122 


.054689 


.154 


.076747 


.091 


.035585 


.123 


.055345 


.155 


.077469 


.092 


.036162 


.124 


.056003 


.156 


.078194 


.093 


.036741 


.125 


.056663 


.157 


.078921 


.094 


.037323 I 


.126 


.057326 


.158 


.079649 


.095 


.037909 1 


.127 


.057991 


.159 


.080380 


.096 


.038496 


.128 


.058658 


.160 


.081112 


.097 


.039087 


.129 


.059327 


.161 


.081846 


.098 


.039680 


.130 


.059999 


.162 


.082582 


.099 


.040276 


.131 


.060672 


.163 


.083320 


.100 


.040875 


.132 


.061348 


.164 


.034059 


.101 


.041476 i 


.133 


.062026 1 


.165 


.084801 



X ^33 ) 



The Areas of the Segments of a Circle. 





'S'e^. Area. 


i 2 


Segc Area. 


h 


Seg. Area. 


>i^ 




>'S: 




1 >'S 




.166 


.085544 


\ .198 


.110226 


! .230 


.136465 


.167 


.086289 


.199 


.111024 


.231 


.137307 


.163 


.087036 


.200 


,111823 


.232 


,138150 


.169 


.087785. 


.201 


.112624 


.233 


.138995 


.170 


.038535 


.202 


.113426 


.234 


.139841 


.171 


.039287 


.203 


.114230 


.235 


.140688 


.172 


.090041 


.204 


.115035 


.236 


.141537 


.173 


.090797 


1 .205 


.115842 


.237 


.142387 


.174 


.091554 


^ .206 


,116650 


1 .238 


.143238 


.175 


.092313 


.207 


.117460 


.239 


.144091 


.176 


.093074 


.208 


.118271 


.240 


.144944 


.177 


.093836 


.209 


.119083 


.241 


.145799 


.178 


.094601 


.210 


.119897 


,242 


.146655 


.179 


.095366 


.211 


.120712 


,243 


.147512 


.180 


.096134 


.212 


.121529 


.244 


.148371 


.181 


.096903 


.213 


.122347 


.245 


.149230 


.182 


.097674 


.214 


.123167 


.246 


.150091 


.183 


.098447 


.215 


.123988 


.247 


.150953 


.184 


.099221 


.216 


.124810 


.248 


.151816 


.185 


.099997 


.217 


.125634 


.249 


.152680 


.186 


.100774 


,218 


.126459 


.250 


.153546 


.187 


,101553 


.219 


.127285 


.251 


.154412 


-.188 


.102334 


.220 


.128113 


.252 


.155280 


.189 


.103116 


.221 


.128942 


.253 


.156149 


.190 


.103900 


.222 


.129773 


.254 


.157019 


.191 


.104685 


.223 


.130605 


.255 


.157890 


.192 


.105472 


.224 


.131438 


.256 


.158762 


.193 


.106261 


.225 


.132272 


.257 


.159636 


.194 


.107051 1 


.226 


.133108 


,258 


.160510 


.195 


.107842 


.227 


.133945 


.259 


.161336 


.196 


.108636 


.228 


.134784 


,260 


.162263 


.197 


.109430 


.229 


.135624 i 


.261 


.163140 



JJ2 



(234) 



The Areas of the Segments of a Circle. 



1. 


Seg. Area. 


Is 


Seg. Area. 




Seg. Area. 


>m 




>-| 




>w 




.262 


.164019 


.294 


.192684 


.326 


.222277 


.263 


.164899 


.295 


.193596 


.327 


.223215 


.261 


.165780 


.296 


.194509 


.328 


.224154 


.265 


.166663 


.297 


.195422 


.329 


.225093 


.266 


.167546 


-.298 


.196337 


.330 


.226033 


.267 


.168430 


.299 


.197252 


.331 


.226974 


.268 


.169315 


.300 


.198168 


•332 


.227915 


.269 


.170202 


.301 


.199085 


.333 


.228858 


.270 


.171089 


.302 


.200003 


.334 


.229801 


.271 


.171978 


.303 


.200922 


.335 


.230745 


.272 


.172867 


.304 


.201841 


.336 


.231689 


.273 


.173758 


.305 


.202761 


.337 


.232634 


.274 


.174649 


.306 


.203683 


.338 


.233580 


.275 


.175542 


.307 


.204605 


.339 


.234526 


.276 


.176435 


.308 


.205527 


.340 


.235473 


.277 


.177330 


.309 


.206451 


.341 


.236421 


.278 


.178225 


.310 


.207376 


.342 


.237369 


.279 


.179122 


.311 


.208301 


.343 


.238318 


.280 


.180019 


.312 


.209227 


.344 


.239268 


.281 


.180918 


.313 


.210154 


.345 


.240218 


.282 


.181817 


.314 


.211082 


.346 


.241169 


.263 


.182718 


.315 


.212011 


.347 


.242121 


.284 


.183619 


.316 


.212940 


.348 


.243074 


.285 


.184521 


.317 


.213871 


.349 


.244026 


.286 


.185425 


.318 


.214802 


.350 


.244980 


.287 


.186329 


.319 


.215733 


.351 


.245934 


.288 


.187234 


.320 


.216666 


.352 


.246889 


.289 


.188140 


.321 


.217599 


.353 


.247845 


.290 


.189047 


.322 


.218533 


.354 


.248801 


.291 


.189955 


.323 


.219468 


.355 


.249757 


.292 


.190864 


.324 


.220404 


.356 


.250715 


.293 


.191775 


.325 


.221340 


.357 


251673 



(235) 



The Areas of the Segments of a Circle. 



1 6 


Seg. Area. 


13 


Seg. Area. 


Is 


Seg. Area. 


.358 


.252631 


.390 


.283592 


.422 


.315016 


.359 


.253590 


.391 


.284568 


.423 


.316004 


.360 


.254550 


.392 


.285544 


.424 


.316992 


.361 


.255510 


.393 


.286521 


.425 


.317981 


.362 


.256471 


.394 


.287498 


.426 


.318970 


.363 


.257433 


.395 


.288476 


.427 


.319959 


.364 


.258395 


.396 


.289453 


•428 


.320948 


.365 


.259357 


.397 


.290432 


.429 


.321938 


.366 


.260320 


.398 


.291411 


.430 


.322928 


.367 


.261284 


.399 


.292390 


.431 


.323918 


.368 


.262248 


.400 


.293369 


.432 


.324909 


.369 


.263213 


.401 


.294349 


.433 


.325900 


.370 


.264178 


.402 


.295330 


.434 


.326892 


.371 


.265144 


.403 


.296311 


.435 


.327882 


.372 


.266111 


.404 


.297292 


.436 


.328874 


.373 


.267078 


.405 


.298273 


.437 


.329866 


.374 


.268045 


.406 


.299255 


.438 


.330858 


.375 


.269013 


.407 


.300238 


.439 


.331850 


.376 


.269982 


.408 


.301220 


.440 


.332843 


.377 


.270951 


.409 


.302203 


.441 


.333836 


.378 


.271920 


.410 


.303187 


.442 


.334829 


.379 


.272890 


.411 


.304171 


.443 


.335822 


.380 


.273861 


.412 


.305155 


.444 


.336816 


.381 


.274832 


.413 


.306140 


.445 


.337810 


.382 


.275803 


.414 


.307125 


.446 


.338804 


.383 


.276775 


.415 


.308110 


.447 


.339798 


.384 


.277748 


.416 


.309095 


.448 


.340793 


.385 


.278721 


.417 


.310081 


.449 


.341787 


.386 


.279694 


.418 


.311068 


.450 


.342782 


.387- 


.280668 


.419 


.312054 


.451 


.343777 


.388 


.281642 


.420 


.313041 


.452 


.344772 


.389 


.2S2617 


.421 


.314029 


.453 


.345768 



( 236 ) 



The Areas of the Segments of a Circle. 



-o 




'§ 




"S 




i ^ 


Seg. Area. 


w c3 


Seg. Area. 


li 


Seg. Area. 


>'^ 




l>i» 




'^m 




.454 


.346764 


.470 


.362717 


.486 


.378701 


.455 


.347759 


.471 


.363715 


.487 


.379700 


.456 


.348755 


.472 


.364713 


.488 


.380700 


.457 


.349752 


.473 


.365712 


.489 


.381699 


.458 


.350748 


.474 


.366710 


.490 


.332699 


.459 


.351745 


.475 


.387709 


.491 


.383699 


.460 


.352741 


.476 


.368708 


.492 


.384699 


.461 


.353739 


.477 


.369707 


.493 


.385699 


.462 


.354736 


.478 


.370706 


.494 


.386699 


.463 


.355732 


.479 


.371705 


.495 


.387699 


.464 


.356730 


.480 


.372764 


.496 


.388699 


.465 


.357727 


.481 


.373703 


.497 


.389699 


.466 


.3.58725 


.482 


.374702 


.498 


.390699 


.467 


.359723 


.483 


..375702 


.499 


.391699 


.468 


.360721 


.484 


.376702 


.500 


.392699 


.469 


.361719 


1 .485 


.377701 


1 





1 



APPENDIX. 



GATJGIlVrG, 



The business of cask-gauging is commonly performed 
by two instruments, namely, the gauging or sliding rule, 
and the gauging or diagonal rod. 

1. OP THE GATJGI?fG RULE. 

This instrument serves to compute the contents of casks, 
&0C. after the dimensions have been taken. It is a square 
rule, having various logarithmic lines on its four sides or 
faces, and three sliding pieces, running in grooves, in three 
of them. 

Upon the first face are three lines, namely, two marked 
A, B, for multiplying and dividing; and the third, MD, for 
malt depth, because it serves to gauge malt. The middle 
one B is on the slider, and is a kind of double line, being 
marked at both the edges of the slider, for applying it to 
both the lines A and MD. These three lines are all of the 
same radius or distance from one to 10, each containing 
twice the length of the radius. A and B are placed and 
numbered exactly alike, each beginning at 1, which may be 
either 1, or 10, or 100, &c. or J, or .01, or .001, &c. but 
whatever it is, the middle division, 10, will be ten times as 
much, and the last division 100 times as much. But 1 on 
the line MD is opposite 215, or more exactly 2150,4 on the 



238 GAUGING. 

other lines, which number 2150,4 denotes the cubic inches 
in a malt bushel; and its divisions numbered retrograde to 
those of A and B. On these two lines are also several 
other marks and letters ; thus, on the line A are MB, for 
malt bushel, at the number 2150.4; and A for ale, at 
282, the cubic inches in an ale gallon ; and on the line B 
is W, for wine, at 231, the cubic inches in a wine gallon; 
also, si, for square inscribed, at ,707, the side of a square 
inscribed in a circle whose diameter is 1 ; se, for square 
equal, at .886, the side of a square which is equal to the 
same circle; and c, for circumference, at 3.1416, the cir- 
cumference of the same circle. 

On the second face, or that opposite the first, are a slider 
and four lines, marked D, C, D, E, at one end, and root, 
square, root, cube, at the other; the lines C and D contain- 
ing respectively the square and cubes of the opposite num- 
bers on the lines D, D ; the radius of D being double to that 
of A, B, C, and triple to that of E ; so that whatever the 
first 1 on D denotes, the first on C is the square of it, and 
the first on E the cube of it ; so if D begin with 1, C and E 
will begin with 1; but if D begin with 10, C will begin 
with 100, and E with 1000; and so on. On the line C are 
marked oc at .0796, for the aroa of the circle, whose cir- 
cumference is 1 ; and od at .7654, for the area of the circle 
whose diameter is 1. Also on the line D, are WG, for 
wine gauge, at 17.15; AG for ale gauge, at 18.95; and 
MR for malt round, at 52.32 ; these three being the gauge 
points for round and circular measure, and are found by 
dividing the square roots of 231, 282, and 2150.4 by the 
square root of .7854 : also, MS, for malt square, are marked 
at 46.37, the malt gauge point for square measure, being 
the square root of 2150.4. 

On the third face are three lines, one. on a slider marked 
N ; and two on the stock, marked SS and SL, for segment 
standing and segment lying, which serve for ullaging stand- 
ing and lying casks. 

And on the fourth, or opposite face, are a scale of inches, 
and three other scales marked spheroid, or 1st variety, 2d 
variety, 3d variety ; the scale for the fourth or conic variety, 



GAUGING. 239 

being on the inside of the slider in the third face. The use 
of these lines is to find the mean diameters of casks. 

Besides all those lines, there are two others on the in- 
sides of the first two sliders, being continued from the one 
slider to the other. The one of these is a scale of inches, 
from 12^ to 36; and the other is a scale of ale gallons, be- 
tween the corresponding numbers 435 and 3.61 ; which 
form a table to show, in ale gallons, the contents of all 
cylinders whose diameters are from 12^ to 36 inches, their 
common altitude being 1 inch. 

The use of the Gauging Rule* 

PROBLEM I. 

To multiply two numbers, as 12 and 25. 

Set 1 on B, to either of the given numbers, as 12, on A ; 
then against 25 on B, stands 300 on A ; which is the pro- 
duct. 

PROBLEM n. 

To divide one number by another, as 300 Ji/ 25. 

Set 1 on B, to 25 on A ; then against 300 on A, stands 
12 on B for the quotient. 

. PROBLEM m. 

To find a fourth proportional, as to 8, 24, and 96. 
Set A on B, to 24 on A ; then against 96 on B, is 288 
on A, the 4th proportional to 8, 24, 96 required. 

PROBLEM IV. 

To extract the square root, as of 225. 

The first one on C standing opposite the one on D, on 
the stock ; then against 225 on C, stands its square root 
15 on D. 

PROBLEM V. 

To extract the cube root, as of 3375. 



240 GAUGING. 

The line D on the slide being set straight with E ; oppo- 
site 3375 on E stands its cube root 15 on D. 

PROBLEM VI. 

To find a mean proportional^ as between 4 and 9. 

Set 4 on C, to the same 4 on D ; then against 9 on C, 
stands the mean proportional 6 on D. 

PROBLEM VIL 

To find numbers in duplicate proportion. 

As, to find a number which shall be to 120, as the square 

of S to the square of 2. 

Set 2 on D, to 120 on C; then against 3 on D, stands 
270 on C; for the answer. 

PROBLEM VIIL 

To find numbers in subduplicate proportion. 

As, to find a number which shall be to 2 as the root of 270 
to the root of 120. 

Set 2 on D, to 120 on C, then against 270 on C, stands 
3 on D, for the answer. 

PROBLEM IX. 

To find the numbers in triplicate proportion. 

As, to find a number which shall be to 100, as the cube of 
36 is to the cube of 40. 

Set 40 on D, to 100 on E ; then against 36 on D, stands 
72.9 on E, for the answer. 

PROBLEM X. 

To find numbers in subtriplicate proportion. 

As, to find a number which shall be to 40, as the cube root 
of 72.9 is to the cube root of 100. 



GAUGING. 241 

Set 40 on D, to 100 on E ; then against 72.9 on E stands 
36 on D, for the answer. 

PROBLEM XL 

To compute malt husliels by the line JMD. 

ASi to find the malt bushels in the couch, fioor, or cistern, 
whose length is 230, breadth 58,2 and depth 5.4 inches. 

Set 230 on B, to 5.4 on MD ; then against 58.2 on A, 
stands 33.6 bushels on B, for the answer. 

Note. — The uses of the other marks on the rule, will ap- 
pear in the examples further on. 

OF THE GAUGIIVG OR DIAGONAL EOD. 

The diagonal rod is a square rule, having four faces; 
being commonly four feet long, and folding together by 
joints. This instrument is used both for gauging and mea- 
suring casks, and computing their contents, and that from 
one dimension only, namely, the diagonal of the cask, or 
the length from the middle of the bung-hole to the meeting 
of the head of the cask with the stave opposite to the bung ; 
being the longest straight line that can be drawn within 
the cask from the middle of the bung. And, accordingly, 
on one face of the rule is a scale of inches for measuring 
this diagonal ; to which are placed the areas, in ale gallons, 
of circles to the corresponding diameters, in like manner 
as the lines on the under sides of the three slides in the 
sliding rule. 

On the opposite face, are two scales of ale and wine 
gallons, expressing the contents of casks having the cor- 
responding diagonals. And these are the lines which 
chiefly form the difference between this instrument and 
the sliding rule: for all their other lines are the same, and 
are to be used in the same manner. 



The rod being applied within the cask at the bung-hole^ 
the diagonal was found to be 34.4 inches; required tl:& 
content in gallons. 
X 



242 GAroiNG. 

Now to 34.4 inches correspond, on the rod, 90| ale gal- 
lons, or 111 wine gallons, the content required. 

Note, — The contents exhibited by the rod, answer to the 
most common form of casks, and fall in between the 2d 
and 3d varieties following. 

OF CASKS AS DIVIDED INTO VARIETIES. 

It is usual to divide casks into four cases or varieties, 
which are judged of from the greater or less apparent cur- 
vature of their sides ; namely, 

1. The middle frustrum of a spheroid, 

2. The middle frustrum of a parabolic spindle, 

3. The two equal frustrums of a paraboloid, 

4. The two equal frustrums of a cone. 

And if the content of any of these be computed in inches, 
by their proper rules, and this be divided by 282, or 231, 
or 2150.4, the quotient will be the content in ale gallons 
or wine gallons, or malt bushels, respectively. Because 

282 cubic inches make 1 ale gallon. 

231 ... 1 wine gallon. 

2150.4 ... 1 malt bushel. 

And the particular rule will be for each as in the follow- 
ing problems : 

PROBLEM Xn. 

To find the content of a cask of the first form. 

To the square of the head diameter add double the 
square of the bung diameter, and multiply the sum by the 
length of the cask. Then let the product 

be multiplied by .0009^, or divided by 1077, for al 
gallons ; 

and multiplied by .0011^, or divided by 882 for wine 
! gallons. 

EXAMPLES. 

1. Required the content of a spheroidal cask, whose 
length is 40, and bung and head diameters 32 and 24 
ijnches. 



24 


82 


GAUGIWO. 






169* 


24 


32 








96 


64 




((fjmf 






48 


96 
1024 


1 


r 




576 











2 












2048 


104966 




104960 




576 


.0009i 




.OOlli 




2624 


944640 




1154560 




40 


26240 




34987 



104960 ale 97.0880 gallons. 118.9547 wine* 
By the Gauging RuUm 

Having set 40 on C, to the ale gauge 32-82 on D* 
against 

24 on D, stands 21.3 on C, 

32 on D, stands 38.0 on C, 

the same 38.0 

sum 97^3 ale gallons. 

And having set 40 on C, to the wine gauge 29.7, on IX 
against 

24 on D, stands 26.1 on C, 

32 Dn D, stands 46.5 on C, 

the same 46.5 

119.1 wine gallons. 



Ex. 2. Required the content of the spheroidal cask, 
whose length is 20, and diameters 12 and 16 inches. 

. 5 12.136 ale gallons, 
^^* I 14,869 wine gallons^ 



244 



PROBLEM Xm. 

To find the content of a cask of the second form. 

To the square of the head diameter, add double the 
square of the bung diameter, and from the sum take f or 
Y*^ of the square of the difference of the diameters ; then 
multiply the remainder by the length, and the product 
again by .0009? for ale gallons, or by .0011^ for wine 
gallons. 

EXA3IPLES. 

1. The length being 40, and diameters 24 and 32, re- 
quired the content. 




32 




24 




8 


2624.0 


8 


25.6 


64 


2598.4 


4 


40 



103936 
.0009i 

935424 
25984 



103936 
.00111 



1143296 
34645 



25.6 



103936 ale 96.1408 gall. 117.9741 wine. 



By the Gauging Rule. 

Having set 40 on C, to 32.82 on D, against 8 on D 
stands 2.4 on C ; the j\ of which is 0.96. This taken from 
the 97.3 in the last form, leaves 96.3 ale gallons. 

And having set 40 on C, to 29.7 on D, against 8 on D, 
stands 2.9 on C ; the y\ of which is 1.16. This taken from 
the 119.1 in the last form, leaves 117.9 wine gallons. 



GATTGIWG. ^45 

^ Ex. 2. Hequired the content when the length is 30, and 

the diameters 12 and 16^ 

I 12.018 ale gallons, 
' 14.724 wine galloni. 



Ans. 



PROBLEM XIV. 

To find the content of a cask of the third form. 

To the square of the bung diameter add the square of the 
head diameter; multiply the Bum by the length, and the 
product again by .0014 for ale gallons, or by .0017 for 
wine gallons. 

EXAMPLES. 

1. Required the content of a cask of the third form, 
when the length is 40, and the diameters 24 and 32. 





By 

to 26. 


^^^^^ 


5_\ 




1024 
676 


64000 
.0014 

256 
64 

ale 89.6 
the Gauging 
8 on D ; then 


Rule. 

against 


64000 
.0017 


1600 
40 


448 
64 


64000 
40 on C, 


108.8 unne. 



24 on D, stands 32.0 on C 
32 on D, stands 57.3 on C 

sum 89,3 ale gallons^ 
And having set 40 on C, to 24.25 on D ; then 
against 24 on D, stands 39.1 on C 
52 on D, stands 69.6 on C 

sum 108.9 wine gallons. 
X2 



246 GAUGING. 

Ex. 2. Required the content when the length is 20, 
and the diameters 12 and 16. 

. 01.2 ale gallons. 
( 13.6 wine gallons. 

PROBLEM XV. 

To find the content of a cask of the fourth form. 
Add the square of the difference of the diameters to three 
times the square of their sum; then multiply the sum by 
the length, and the product again by .000231 for ale gallons, 
01- by .00028i for wine gallons. 

EXAMPLES. 

1. Required the content, when the length is 40, and the 
diameters 24 and 32 indies. 



56 


8 


56 


8 


336 


64 


280 


9408 


3136 


9472 


3 


40 




378880 
.000231 

1136640 
7577 60 
75776 



378880 
.000281 

3031040 
757760 
126293 



9408 378880 oZe 87.90016 ^aZZ. 1 07.34933 ii«ne. 

BY THE SLrDIXG RULE. 

Set 40 on C, to 65.64 on D ; 'then against 
8 on D, stands 0.6 on C 
56 on D, stands 29.1 on C 
29.1 
29.1 



sum 87.9 ale gallons. 



GArGING. 247 

And set 40 on C, to 59.41 on D ; then against 

8 on D, stands 0.7 

66 on D, stands 35.6 

35.6 

35.6 

sum 107.5 wine gall. 

Ex. 2. What is the content of a conical cask, the length 
being 20, and the bung and head diameters 16 and 12 
inches ? 



. i 10.985 ale gallons. 
( 13.418 wine gallons. 



PROBLEM XVI. 

To find the content of a cask by four dimensions. 

Add together the squares of the bung and head diameters, 
and the square of double the diameter taken in the middle 
between the bung and head : then multiply the sum by the 
length of the cask, and the product again by .0004| for ale 
gallons, or by .0005| for wine gallons. 

EXAMPLES. 

1. Required the content of any cask whose length is 40, 
the bung diameter being 32, the head diameter 24, and 
the middle diameter between the bung and the head 28^ 
inches. 

57.5 24 32 

57.5 24 32 



2875 


96 


64 


4025 


48 


96 


2875 







3306.25 
1024 
676 

4906.25 



676 1024 



248 



4906.25 
40 




196250 
.00041 


196250 
.00051 


785000 
130833 


981250 
130833 



ale 91.5833 gallons. 111.2083 wine. 

BY THE SLIDING RULE. 

Set 40 on C, to 46.4 on D ; then against 
24 on D, stands 10.5 
32 on D, stands 19.0 
67^ on D, stands 62.0 

sum 91.5 ale gallons. 
Set 40 on C, to 24.0 on D; then against 
24 on D, stands 13.0 
32 on D, stands 23.2 
67^ on D, stands 75.0 

sum 111.2 wine gallons. 
Ex. 2. What is the content of a cask, whose length is 20, 
the bung diameter being 16, the head diameter 12, and the 
diameter in the middle between them 14f ? 

. i 11.4479 ale gallons, 
^^^* I 13.9010 wine gallons. 

PROBLEM XVII. 

To find the content of any cask from three dimensions only. 

Add into one sum 39 times the square of the bung di- 
ameter, 25 times the square of the head diameter, and 
26 times the product of the two diameters : then multi- 
ply the sum by the length, and the product again by 
,00034 .00034 

— g— for wine gallons, or by — tt — or .00003Jj» for ab 

gallons. 



249 



EXAMPLES. 



1. Required the content of a cask, whose length is 40, 
and the bung and head diameters 32 and 24. 

32 24 32 

32 24 24 



64 
96 

1024 

39 

9216 
.072 


96 

48 

576 
25 

2880 
1152 


128 
64 

768 
26 

4608 
1536 




39936 


14400 
39936 
19968 


19668 












74304 
40 






2972160 
.00034 


2972160 

.oooos^-v 




11888640 
8916480 




8916480 
270196 


Q'il 


010.53440 


VJl 


91.86676 ale gall 



112.2816 wine gall. 
Ex. 2, What is the content of a cask, whose length is 
20, and the bung and head diameters 16 and 12 ? 

^^g i 11.4833 ale gallons. 
( 14.0352 wine gallons. 

Note. — This is the most exact rule of any, for three di- 
mensions only ; and agrees nearly with the diagonal rod. 



250 



OF THE ULLAGE OF CASKS. 



The ullage of a cask is what it contains when only partly 
filled. And it is considered in two positions, namely, as 
standing on its end with the axes perpendicular to the ho- 
rizon, or as lying on its side with the axes parallel to the 
horizon. 

PROBLEM XVIII. 
To find the ullage by the Sliding Rule, 

By one of the preceding problems find the whole content 
of the cask. Then set the length on N, to 100 on SS, for 
a segment standing, or set the bung diameter on N, to 100 
on SL, for a segment lying ; then against the wet inches on 
N, is a number on SS or SL, to be reserved. 

Next, set 100 on B, to the reserved number on A; then 
against the whole content on B, will be found the ullage on 
A. 

EXAMPLES. 

1. Required the ullage answering to 10 wet inches of a 
standing cask, the whole content of which is 92 gallons, 
and length 40 inches. 

Having set 40 on N, to 100 on SS; then against 10 on 
N, is 23 on SS, the reserved number. 

Then set 100 on B, to 23 on A ; and 

against 92 on B, is 21.2 on A, the ullage required. 

Ex. 2. What is the ullage of a standing cask, whose 
whole length is 20 inches, and content 11^ gallons; the 
wet inches being 5 ? Ans. 2.65 galls. 

Ex. 3. The content of a cask being 92 gallons, and the 
bung diameter 32, required the ullage of the segment lying 
when the wet inches are 8. Ans. 16.4 galls 

PROBLEM XIX. 
To ullage a standing cask by the pen. 
Add all together, the square of the diameter at the «Uf- 



GAUOrNG. 251 

face of the liquor, the square of the diameter of the nearest 
end, and the square of double the diameter taken in the 
middle between the other two ; then multiply the sum by 
the length between the surface and nearest end, and the 
product again by .0004| for ale gallons, or by .0005| for 
wine gallons, in the less part of the cask, whether empty or 
filled. 



The three diameters being 24, 27, and 29 inches, re- 
quired the ullage for 10 wet inches. 



24 


29 


54 




24 


29 


54 


2916 


— 






841 


96 


261 


216 


576 


48 


68 


270 












4333 


576 


841 


2916 


10 




43330 




43330 




.00041 




.00051 




173320 




216650 




28885 


)ns. 


28885 


ale 20.2205 gall< 


24.5535 wine. 



PROBLEM XX. 

To ullage a lying cash hy the 'pen. 

Divide the wet inches by the bung diameter, find the 
quotient in the colunm of versed sines, in the table of cir- 
cular segments at page 231 of the book, taking out its cor- 
responding segment. Then multiply this segment by the 
whole content of the cask, and the product again by 1^ for 
the ullage required, nearly. 



252 GAUGING. 

EXAMPLE I. 

Supposing the bung diameter 32, and content 92 ale gal- 
Ions ; to find the ullage for 8 wet inches. 

32)8(.25, whose tab. seg. is .153546 
92 



307092 
1381914 

14.126-232 
^ is 3.531558 

17.657790 Ans. 



EXAMPLE II. 

Taking the length of the cask 40, bung diameter 32, 
head diameter 24 ; and supposing the wet inches to be 8. 
What is the ullage ? 

Ans. 18 ale gallons. 

Of Guaging Casks by their Mean Diameters. 

PROBLEM I. 

To find the Mean Diameter of a Cask (^f any of the four 
varieties, having given the bung and head diameters. 

Divide the head diameter by the bung diameter, and find 
the quotient in the first column of the following table, 
marked Qu. Then if the bung diameter be multiplied by 
the number on the same line with it, and in the column an- 
swering to the proper variety, the product will be the true 
mean diameter, or the diameter of a cylin ler of the same 
content with the cask proposed, cutting oflf four figures for 
decimals. 



253 



au 


L Var. 


2 Var. 


3 Var 


4 Var. 


au 


1 Var. 


2 Var. 


3 Var. 


4 Var. 


50 


8660 


8465 


7905 


7637 


76 


9270 


9227 


8881 


8827 


51 


8680 


8493 


7937 


7681 


77 


9296 


9258 


8944 


8874 


52 


8700 


8520 


7970 


7725 


78 


9324 


9290 


8967 


8922 


53 


8720 


8548 


8002 


7769 


79 


9352 


9320 


9011 


8970 


54 


8740 


8576 


8036 


7813 


80 


9380 


9352 


9055 


9018 


55 


8760 


8605 


8070 


7858 


81 


9409 


9383 


9100 


9066 


56 


8781 


8633 


8104 


7902 


82 


9438 


9415 


9144 


9114 


57 


8802 


8662 


8140 


7947 


83 


9467 


9446 


9189 


9163 


58 


8824 


8690 


8174 


7992 


84 


9496 


9478 


9234 


9211 


59 


8846 


8720 


8210 


8037 


85 


9526 


9510 


9280 


9260 


60 


8869 


8748 


8246 


8082 


86 


9556 


9542 


9326 


9308 


61 


8892 


8777 


8282 


8128 


87 


9586 


9574 


9372 


9357 


62 


8915 


8806 


8320 


8173 


88 


9616 


9606 


9419 


9406 


63 


8938 


8835 


8357 


8220 


89 


9647 


9638 


9466 


9455 


64 


8962 


8865 


8395 


8265 


90 


9678 


9671 


9513 


9504 


65 


8986 


8894 


8433 


8311 


91 


9710 


9703 


9560 


9553 


66 


9010 


8924 


8472 


8357 


92 


9740 


9736 


9608 


9602 


67 


9034 


8954 


3511 


8404 


93 


9772 


9768 


9656 


9652 


68 


9060 


8983 


8551 


8450 


94 


9804 


9801 


9704 


9701 


69 


9084 


9013 


8590 


8497 


95 


9836 


9834 


9753 


9751 


70 


9110 


9044 


8631 


8544 


96 


9868 


9867 


9802 


9800 


71 


9136 


9074 


8672 


8590 


97 


9901 


9900 


9851 


9850 


72 


9162 


9104 


8713 


8637 


98 


9933 


9933 


9900 


9900 


73 


9188 


9135 


8754 


8685 


99 


9966 


9966 


9950 


9950 


74 


9215 


9166 


8796 


8732 


100 


10000 


10000 


10000 


10000 


75 


9242 


9196 


8838 


8780 













Supposing the diameters to be 32 and 24, it is required 
to find the mean diameter for each variety. 

Dividing 24 by 32, we obtain .75; which being found in 
the column of quotients, opposite thereto stand the numbers 

f .9242"! which being each f 29.5744"! ^^^ ^^^ corres- 

! .9196 ! miiUlnliRd hv 39_ ! 29.4272 ! nondino- r 



.9196 [multiplied by 32, 
I .8838 f produce respect- 
L. 8780 J ively 
Y 



29.4272 I ponding mean 
28.2816 f diameters re- 
28.0960 j quired. 



254 GAUGING. 

BY THE SLIDING RULE. 

Find the difference between the bung and head diame- 
ters on the fourth face of the rule, or inside of the third 
slider; and opposite thereto is, for each variety, a number 
to be added to the head diameter, for the mean diameter 
required. 

So, in the above example, against 8, the difference of 
the diameters, are found the numbers 
5.60 "j u- L, u • f 29.60^ for the respective mean di- 
5.10 I ^jTj . L"f 1 29.10 • ameters; all of which are 

^•^^ I there result ^^'^^ ' ^^^ S'^'*' ^^^^P* '^^ '^^ 
4.12 j ^^^"^^ ^^^^^^ L 28.12 J cond, which is too little. 

So that this method does not give the true mean di- 
ameter. 

PROBLEM ir. 

To find the content of a cask by the mean diameter on the 
Sliding Rule. 
Set the length on C, to the gauge point, 18.95 for ale, 
or 17.16 for wine, on D; then against the mean diameter 
on D, is the content on C. 

EXAMPLE. 

If the bung diameter be 32, the head 24; and the length 
40 inches. 

Having found the mean diameters, as in the last problem, 
and set 40 on C, to 18.95 or 17.15 on D, 
^ f29.57^.S2 f97.4^ f 1 19.5^ on C, as near as can 
c J 29.43 I -r 1 96.5 ' ^ J 118.0 'be judged; which 
^ ) 28.28 [ ^ \ 89.1 f ® I 108.8 (agree nearly with 
"" L28.10j o |^88.0J Ll07.3jthe contents com- 
puled in the preceding chapter. 

SCHOLIUM. 

Having delivered the necessary rules for measuring 
casks, &,c., I do not suppose that any thing more of the 
subject of gauging is wanted to be given in this book. For, 
as to cisterns, couches, &;c. tuns, coolers, &c. coppers, 



TONNAGE. 255 

Stills, &;c. which are first supposed to be in the form of 
some of the solids in the former parts of this work, and 
then measured accordingly, no person can be at a loss con- 
cerning them, who knows any thing of such solids in gene- 
ral; and to treat of them here, would induce me to a long 
and tedious repetition, only for the sake of pointing out the 
proper multipliers or divisors; which is, I think, a reason 
very inadequate to so cumbersome an increase of the book. 

I shall only just observe, that when tuns, &c. of oval 
bases are to be gauged; as those bases really measure 
more than true ellipses of the same length and breadth, 
they ought to be measured by the equi-distant ordinate me- 
thod. 

And that when casks are met with which have different 
head diameters, they may be deemed incomplete casks, 
and their contents considered and measured as the ullage 
of a cask. 

TO FIND THE TONNAGE OF A SKIP. 

The length is taken in a straight line along the rabbet 
of the keel, from the back of the main stern-post to a per- 
pendicular from the fore part of the main stem, under the 
bowsprit, from which subtract -f of the breadth, the re- 
mainder is the length. The breadth is taken at the broad- 
est part of the ship, from the outside to the outside. 

Rule. — Multiply the square of the breadth by the 
length, and divide the product by 188, the quotient will be 
the tonnage. 

Ex. 1. — Required the tonnage of a ship, of which the 
length is 75 feet, and the breadth 26 feet. 

Ans. 26X26X75-M 88=270 tons, nearly, 

Ex. 2.— Length 96, and breadth 33 feet? Ans. 656 tons. 

Note. — This rule is very erroneous, and no other gene- 
ral rule can be given which is perfectly accurate; the best 
way is to find the quantity of water displaced by the ship 
when she is loaded; but as this must be done by means of 
ordinates, the operation is laborious. It is easier to load 
her with ballast, weighing the load as it is put on board. 



256 FALLING BODIES. 

Tlie following rule is a near approximation for ships of 
burden. 

Take the length of the lower deck, from the rabbet of 
the stem to that of the stern-post, and from it subtract 3^2 
of it, for the length. Take the extreme breadth from 
outside to outside, and add it to the length of the lower 
deck, /j of the sum is the depth. Set up this depth from 
the limber strake, where the extreme breadth was taken, 
and at this height take a breadth from outside to outside, 
take another breadth at | of this height, and a third at i 
of the height, add these three to the exireme breadth, 
and i of the sum is the mean breadth. Multiply the 
length, breadth, and depth, and divide three times the 
product by 110 for the tonnage. 



FALLING BODIES. 

The motion described by bodies freely descending by 
their own gravity is, viz: — The velocities are as the times, 
and the spaces as the squares of the times. Therefore, if 
the times be as the numbers ... 1 2 3 4 &:.c. 
The velocities will be also as ... 1 2 .3 4 &.c. 
The spaces as their squares ... 1 4916 &c. 
and the spaces for each time, as . . 1 3 5 7 &c. 
namely, as the series of the odd numbers, which are the 
differences of the squares, denoting the whole spaces; so 
that if the first series of numbers be seconds of time : 

i.e 1" 2" 3" &,c. 

Velocities in feet will be . . . 32^ 64i 96^ &C. 
Spaces in the whole time will be 16^^ 64^ 144^ &c. 
Spaces for each second will be 16^2" '^^i SO-j^g &c. 



FALLING BODIES. 



257 



The following 


able shows the Spaces fallen 


through, and the Velo- 




cities acquired at the end of each 20 seconds. 


Time 
in 

Seconds. 


SPACE. 


VELOCITY. 1 


Each 
Time. 


As tlie 

Squares 

of the Time. 


Fallen 

through in 

Feet and Inches. 


As the 
Times. 


Acquired 

in 

Feet and Inches. 


1 


1 


1 


16 1 


1 


32 2 


2 


3 


4 


64 4 


2 


64 4 


3 


5 


9 


144 9 


3 


96 6 


4 


7 


16 


257 4 


4 


128 8 


5 


9 


25 


402 1 


5 


160 10 


6 


11 


36 


579 


6 


193 


7 


13 


49 


788 1 


7 


225 2 


8 


15 


64 


1029 4 


8 


257 4 


9 


17 


81 


1302 9 


9 


£89 6 


10 


19 


100 


1608 4 


10 


321 8 


11 


21 


121 


1946 1 


11 


353 10 


12 


23 


144 


2316 


12 


386 


13 


25 


169 


2718 1 


13 


418 2 


14 


27 


196 


3152 4 


14 


450 4 


15 


29 


225 


3618 9 


15 


482 6 


16 


31 


256 


4117 4 


16 


514 8 


17 


33 


289 


4648 1 


17 


546 10 


18 


35 


324 


5211 


18 


579 


19 


37 


361 


6806 1 


19 


611 2 


20 


39 


400 


6433 4 


20 


643 4 



EXAMPLE I. 

To find the space descended by a body in 1" and the 
velocity acquired. 

16 1 X 49= 788 1 of space. 

32 2 X 7" = 225 2 of velocity. 
Look into the table at 7" and you have the answers. 

EXAMPLE II. 

To find the time of generating a velocity of 100 feet 
per second, and the whole space descended. 
100 X 12 ^„^, . 

= 3 VqV time. 

32 2 X 12 '^^ 



3"-2j_ ^ ]00 

— ^-^ = 155fV^3- space descended. 



Y2 



258 PENDtLUJI. 

EXAMPLE IV. 

To find the time of descending 400 feet, and the velo- 
city at the end of that time. 

v/400 X 12 ,„ ^^^ . 

== 4' 987 time. 

v/16. 1 X 1 2 

400 X ?^ 169.662 velocity. 



I 



4" 987 

Or these answers can be found from the Table by Pro- 
portion. 



PENDULUM. 

The vibrations of pendulums are as the square roots of 
their lengths ; and as it has been found by many accurate 
experiments, that the pendulum vibrating seconds in the 
latitude of London, is 391 inches long nearly, the length 
of any other pendulum may be found by the following rule, 
viz. — As the number of vibrations given is to 60, so is the 
square root of the length of the pendulum that vibrates 
seconds, to the square root of the length of the pendulum 
that will oscillate the given number of vibrations ; or, as 
the square root of the length of the pendulum given, is to 
the square root of the length of the pendulum that vibrates 
seconds, so is 60 to the number of vibrations of the given 
pendulum. 

Since the pendulum that vibrates seconds, or 60, is 39| 
inches long, the calculation is rendered simple ; for ■\/39| 
X 60=375, a constant number, therefore 375, divided by 
the square root of the pendulum's length, gives the vibra- 
tions per minute, and divided by the vibrations per minute, 
gives the square root of the length of the pendulum. 

EXAMPLE I. 

How many vibrations will a pendulum of 49 inches long 
make in a minute ? 



MECHANICAL POWERS. 259 

376 

—==534 vibrations in a a minute. 
-v/49 

EXAMPLE II. 

What length of a pendulum will it require to make 90 
vibrations in a minute ? 

375 

— QQ =4.16, and 4.16^=17.3056 inches long. 

EXAMPLE III. 

What is the length of a pendulum, whose vibrations will 
be the same number as the inches in its length ? 
V ^(375)^=52 inches long, and 52 vibrations. 
It is proposed to determine the length of a pendulum 
vibrating seconds, in the latitude of London, where a 
heavy body falls through 16^2 feet in the first second of 
time'? 

3.1416 circumference, the diameter being 1. 
16^2 feet=193 inches fall in the 1" of time. 
19^ X 2 =386.QOQ0QO0O _ 
3.1416^ = 9.86965056-'^^'^"^'"''*'^'' 
or 39.11 inches. 
By experiment this length is found to be 39|- inches. 

What is the length of a pendulum vibrating in 2 se- 
conds, and another in half a second ? 
^39i=6.25 X 60=375. 

375 

=12.5 squared = 156.25 inches the length of a 2 se- 

^ [conds' pendulum. 

375 

-—=3.125 squared=9. 765625 inches the length of a i 
120 o 2 

second's pendulum. 



MECHANICAL POWERS, &c. 

The Science of Mechanics is simply the application of 
Weight and Power, or Force and Resistance. The 
weight is the resistance to be overcome; the power is the 
force requisite to overcome that resistance. When the 



260 LEVER. 

force is equal to the resistance, they are in a state of equi- 
librium, and no motion can take place; but when the 
force becomes greater than the resistance, they are not 
in a state of equilibrium, and motion takes place; conse- 
quently, the greater the force is to the resistance, the 
greater is the motion or velocity. 

The Science of Equilibrium is called Statics; the Sci- 
ence of Motion is called Dynamics. 

Mechanical Powers are the most simple of mechanical 
applications to increase force and overcome resistance. 
They are usually accounted six in number, viz. The Le- 
ver — The Wheel and Axle^ — The Pulley ^ — The Inclined 
Plane ^- — The Wedge^ — and the Screw. 

LEVER. 

To make the principle easily understood, we must sup- 
pose the lever an inflexible rod without weight; when this 
is done, the rule to find the equilibrium between the power 
and the weight, is, — Multiply the weight by its distance 
from the fulcrum, prop, or centre of motion, and the power 
by its distance from the same point: if the products are 
equal, the weight and power are in equilibrio, if not, they 
are to each other as their products. 

EXAMPLE I. 

A weight of 100 lbs. on one end of a lever, is 6 inches 
from the prop, and the weight of 20 lbs. at the other end, 
is 25 inches from the prop — What additional weight must 
be added to the 20 lbs. to make it balance the 100 lbs.? 

=24 — 20=4 lbs. weight to be added. 

23 

EXAMPLE II. 

A block of 960 lbs. is to be lifted by a lever 30 feet 
long, and the power to be applied is 60 lbs. — on what 
part of the lever must the fulcrum be placed? 

— — = 16, that is, the weight is to the power as 16 is to 
^^ 30 

1; therefore the whole length— — -= 1|^, the distance 



WHEEL AND AXiLE. 26 1 

from the block, and 30 — lJ-^=28y\, the distance from 
the power. 

EXAMPLE III. 

A beam 32 feet long, and supported at both ends, bears 
a weight of 6 tons, 12 feet from one end, — What propor- 
tion of weight does each of the supports bear? 

12x6 

=2i tons, support at end farthest from the 

32 . 

weight. 

20 X 6 

— — — =3| tons, support at end nearest the weight. 

EXAMPLE IV. 

A beam supported at both ends, and 16 feet long, car- 
ries a weight of 6 tons, 3 feet from one end, and another 
weight of 4 tons, 2 feet from the other end. What pro- 
portion of weight does each of the supports bear? 

3x6 14 X 4 74 

— - — j -— ===--=4i^ tons, end at the 4 tons. 

16 ^ 16 16 1^ 

2x4 13X6 86 ^ , . , r. 

—^ — I j^=— = 5^ tons, end at the 6 tons. 

WHEEL AND AXLE. 

The nature of this machine is suggested by its name. 
To it may be referred all turning or wheel machines of 
different radii: as well-rollers and handles, Cranes, Cap- 
stans, Windlasses, &c. 

The mechanical property is the same as in the lever: 
that is, the product of the weight into the distance at which 
it acts is equal to the product of the power into the dis- 
tance at which it acts, the distances being estimated in di- 
rections perpendicular to those, in which the weight and 
power act respectively, because the wheel and axle is only 
a kind of perpetual lever. 

And hence also this property: The product of the power 
applied, multiplied by its velocity^ is equal to that of the 
iveight to be raised into its velocity. 

When a series of wheels and axles act upon each other, 



262 WHEEL AND AXLE. 

SO as to transmit and accumulate a mechanical advantage, 
whether the communication be by means of cords and belts, 
or of teeth and pinions, the weight will he to the power, as 
the continual product of the radii of the wheels to the con- 
tinual product of the radii of the axles. 

Thus, if the radii of the axles a, 6, c, (Z, e, be each 3 
inches, while the radii of the wheels A, B, C, D, E, be 9, 
6, 9, 10 and 12 inches, respectively: then W : P :: 9 X 6 
X9x 10X12:3x3x3x3x 3, or as 240:1. A com- 
putation, however, in which the effect of friction is disre- 
garded. 

A train of wheels and pinions may also serve for the 
augmentation of velocities. Thus, in the preceding exam- 
ple, whatever motion be given to the circumference of 
the axle e, the rim of the wheel A will move 240 times as 
fast. 

And if a series of 6 wheels and axles, each having their 
diameters in the ratio of 10 to 1 were employed to accumu- 
late velocity, the produced would be to the producing ve- 
locity as 106 to J. that is, as 1,000,000 to 1. 

Note. — A man's power producing the greatest effect, is 
31 lbs. at a velocity of 2 feet per second, or 120 feet per 
minute. 

The Rule to find the power of Cranes is^ viz. 
Divide the product of the driven by the product of the 
drivers, and the quotient is the relative velocity, as 1 : c, 
which multiplied by the length of winch, and by the power 
applied (in lbs.) and divided by the radius of the barrel, 
the quotient will be the weight raised. 

EXAMPLE I. 

A weight of 94 tons is to be raised 360 feet in 15 mi- 
nutes, by a power, the velocity of which is 220 feet per 
minute: — VVhat is the power required? 

=24 feet per minute, velocity of weight 

24X94=2256 ,^^^,^, . , 
-— =:10.2o45 tons power required. 

22U 



PULLEY. 263 

EXAltfPLE II. 

A stone weighing 986 lbs. is required to be lifted: 

What power must be applied, when the power is to the 

weight as 9 is to 2? 

986X2 1972 ^,^,^ 

r= =219f tons power. 

EXAMPLE III. 

A power of 18 lbs. is applied to the winch of a crane, 
the length of which is 8 inches; the pinion makes 12 re- 
volutions for 1 of the wheel, and the barrel is 6 inches 
diameter. 

8 X 2 X 22 

=50.28 circumferenceof the winch's circle. 

7 

50.28 X 12=603.36 inches velocity of power on winch 

to 1 revolution of the barrel. 

603.36 X 18 = 10860.48 . 

=571.604 lbs. w^eiffht, 

6 X 22 ^ 

— -—=18.857 . . 19. 

that can be raised by a power of 18 lbs. on this crane. 

PULLEY. 

There are two kinds of Pulleys, the Jlxed and the move- 
able. From the fixed Pulley no power is derived; it is as 
a common beam used in weighing goods, having the two 
ends of equal weight, and at the same distance from the 
centre of motion; the only advantage gained by the fixed 
pulley, is in changiug the direction of the power. 

From the moveable pulley power is gained; it operates 
as a lever of the second order ; for if one end of a string 
be fixed to an immoveable stud, and the other end to a 
moveable power, the string doubled and the ends parallel, 
the pulley that hangs between is a lever ; the fixed end of 
the string being the fulcrum, and the other the moveable 
end of the lever : hence the power is double the distance 
from the fulcrum, than is the weight hung at the pulley ; 
and therefore the power is to the weight as 2 is to 1. This 



264 INCLINED PLANE. 

is all the advantage gained by one moveable pulley, for 
two, twice the advantage ; for three, thrice the advantage; 
and so on for every additional moveable pulley. 

From this the following rule is derived : — Divide the 
weight to be raised by twice the number of moveable pul- 
leys, and the quotient is the power required to raise the 
weight. 

EXAMPLE I. 

What power is requisite to lift 100 lbs. when two blocks, 
of three pulleys or sheives each, are applied, the one block 
moveable and the other fixed ? 

— = 16| lbs. the power required, 3 sheives X 2 = 6. 
6 

EXAMPLE II. 

What weight will a power of 80 lbs. lift, when applied 
to a 4 and 5 shelved block and tackle, the 4 shelved block 
being moveable ? 

80 X 8 = 640 lbs. weight raised. 

INCLINED PLANE. 

When a body is drawn up a vertical plane, the whole 
weight of the body is sustained by the power that draws 
or lifts it up : hence the power is equal to the weight. 

When a body is drawn along an horizontal (truly level) 
plane, it takes no power to draw it, (save the friction oc- 
casioned by the rubbing along the plane.) 

From these two hypotheses, if a body is drawn up an 
inclined plane, the power required to raise it is as the in- 
clination of the plane ; and hence when the power acts 
parallel to the plane, the length of the plane is to the 
weight, as the height of the plane is to the power ; for the 
greater the angle, the greater the height. 

EXAMPLE I. 

What power is requisite to move a weight of 100 lbs. 
up an inclined plane, 6 feet long and 4 feet high ? 
If6:4 :: 100 : 66| lbs. power. 



WEDGE AND SCEEW. 266 

EXAMPLE II. 

A power of 68 lbs. at the rate of 200 feet per minute, 
is applied to pull a weight up an inclined plane, at the 
rate of 50 feet per minute — When the plane is 37 feet 
long and 12 feet high, how much will be the weight drawn ? 
As 12 : 37 :: 68 X 200 : 50 X 838| 

68 X 200 X 37 503200 ^^^^ ,, . . 

= - — — - = 838f lbs. weiffht. 

12 X 60 600 " *= 

WEDGE. 

The Wedge is a double inclined plane, and therefore 
subject to the same rules ; or the following rule, which is 
particularly for the wedge, but drawn from its near con- 
nection to the inclined plane, is, — If the power acts per- 
pendicularly upon the head of the wedge, the power is to 
the pressure which it exerts perpendicularly on each side 
of the wedge, as the head of the wedge is to its side: 
hence, it is evident, that the sliarper or thinner the wedge 
is, the greater will be the power. 

But the power of the wedge being not directly accord- 
ing to its length and thickness, but to the length and 
width of the split, or rift, in the wood to be cleft, the rule 
therefore is of little use in practice ; besides, the wedoce 
is very seldom used as a power ; for these reasons, the 
nature of its properties and effects need not be here dis- 
cussed. 

SCKEW\ 

The screw is a cord wound in a spiral direction round 
the periphery of a cyhnder, and is therefore an inclined 
plane, the length being the circumference of the cylinder, 
and the height, the distance between two consecutive 
cords, or threads of the screw, hence, the rule is derived; 
■ — As the circumference of the screw is to the pitch, Or 
distance between the threads, so is the weight to the 
power. 

When the screw turns, the cord or thread runs in a con- 
Z 



266 VELOCITY OF WHEELS. 

tinued ascending line round the centre of the cylinder, and 
the greater the radius of the cylinder, the greater will be 
the length of the plane to its height, consequently, the 
greater the power. A lever fixed to the end of the screw 
will act as one of the second order, and the power gained 
will be as its length, to the radius of the cylinder ; or the 
circumference of the circle described by it, to the circum- 
ference of the cylinder ; hence, an addition to the rule is 
produced, which is, — If a lever is used, the circumference 
of the lever is taken for, or instead of, the circumference 
of the screw. 

EXAMPLE I. 

What is the power requisite to raise a weight of 8000 
lbs. by a screw of 12 inches circumference and 1 inch 
pitch ? 

As 12 : 1 :: 8000 : 66Gj\ lbs. = power at the circum- 
ference of the screw. 

EXAMPLE II. 

How much would be the power if a lever of 30 inches 
was applied to the screw ? 

Circumference of 30 inches = I884. 

As 188A : 1 :: 8000 : 42,-^0- ^^s. = power with a lever 
of 30 inches long. 



VELOCITY OF WHEELS. 

Wheels are for conveying motion to the different parts 
of a machine, at the same, or at a greater or less velocity, 
as may be required. When two wheels are in motion their 
teeth act on one another alternately, and consequently, if 
one of these wheels has 40 teeth, and the other 20 teeth, 
the one with 20 will turn twice upon its axis for one revo- 
lution of the wheel with 40 teeth. From this the rule is 
taken, which is, — As the velocity required is to the nura- 



VELOCITY OF WHEELS. 267 

ber of teeth in the driver, so is the velocity of the driver 
to ths number of teeth in the driven. 

Note, To find the prop? rlion that the velocities of the 
wheels in a train should bear to one another, subtract the 
less velocity from the greater, and divide the remainder 
by the number of one less than the wheels in the train ; 
the quotient will be the number rising in arithmetical pro- 
gression, from the least to the greatest velocity of the train 
of wheels. 

EXAMPLE I. 

What is the number of teeth in each of three wheels to 
produce 17 revolutions per minute, the driver having 107 
teeth, and making three revolutions per minute ? 

,""— -=7, therefore 3 10 17 are the velocities of 
3 — 1= 2 

the three wheels. 

107X3 

f 10 : 107 : : 3 : 32= =32 teeth. 



By the rule 



17 : 32 ; : 10 : 19=: 



10 
32X 10 



17 



19 teeth. 



EXAMPLE II. 

What is the number of teeih in each of 7 wheels, to pro- 
duce 1 revolution per minute, the driver having 25 teeth, 
and making oQ revolutions per minute ? 

„ , -^=9, therefore 56 46 37 28 \^ 10 1, are the 
7 — 1= 6 

progressional velocities. 

46 : 2o : : .56 : 30 Teeth. 

37 : 30 : : 46 : 37 

28 : 37 : : 37 : 49 

19 : 49 ; : 28 : 72 

10 : 72 : : 19 : 137 

1 : 137 ; : 10 : 1370 

It will be observed that the last wheel, in the foregoing 

example, is of a size too great for application ,* to obviate 



268 STEAM ENGINE. 

this difliculty, which frequently arises in tliis kind of train- 
ing, wheels and pinions are used, which give a great conn- 
mand of velocity. — Suppose the velocities of last example, 
and the train only of 2 wheels and 2 pinions. 

gg I = 55 

-— = 18, therefore 56 19 1, are the progres- 

sional velocities. 

19 : 25 : : 56 : 74=teeth in the wheel driven by the 
first driver, and 1 : 10 : : 19 : 190 = teeth, in the se- 
cond driven wheel, 10 teeth being in the drivinu pinion. 
25 drivers 74 driven. 
10 190 ^ 



STEAM ENGINE, 

Boilers — are of various forms, but the most general is 
proportioned as follows, viz. widtli 1, depth 1.1, length 2.5; 
their capacity being, for the most part, two horse more 
than the power of the engine for which they are intended. 

Boulton and Watt allow 25 cubic feet of space for each 
horse power, some of the other engineers allow 5 feet of 
surface of water. 

Steam — arising from water at the boiling point, is equal 
to the pressure of the atmosphere, which is in round num- 
bers, 15 lbs. on the square inch; but to allow for a con- 
stant and uniform supply of steam to the engine, the safety 
valve of the boiler is loaded with three lbs. on each square 
inch. 

The following table exhibits the expansive force of 
steam, expressing the degrees of heat at each lb. of pres- 
sure on the safety valve. 



STEAM ENGINE. 



269 



Degrees of 


Lbs. of 


Degrees of 
Heat. 


Lbs. of 


Degrees of 


Lbs. of 


Heat. 


Pressure. 


Pressure. 


Heat. 


Pressure. 


212° 





268° 


24 


! 298° 


48 


216 


1 


270 


25 


1 299 


49 


219 


2 


271 


26 


300 


60 


222 


3 


273 


27 


301 


61 


225 


' 4 


274 


28 


302 


52 


229 


5 


275 


29 


303 


63 


232 


6 


277 


30 


304 


54 


2.34 


7 


278 


31 


305 


55 


236 


8 


279 


32 


306 


56 


239 


9 


281 


33 


307 


57 


241 


10 


282 


34 


308 


58 


244 


11 


283 


35 


309 


59 


246 


12 


285 


36 


310 


60 


248 


13 


286 


37 


311 


61 


250 


14 


287 


38 


312 


62 


252 


15 


288 


39 


313 


63 


254 


16 


289 


40 


313i 


64 


256 


17 


290 


41 


314 


65 


258 


18 


291 


42 


315 i 


66 


260 


19 


293 


43 


316 


67 


261 


20 


294 


44 


317 


68 


263 


21 


295 


45 


318 


69 


265 


22 


296 


46 


319 ' 


70 


267 


23 


297 


47 


320 ! 


71 



By the following rule the quantity of steam required 
to raise a given quantity of water to any given tempera- 
ture is found. 

Rule. Multiply the water to be warnned by the differ- 
ence of temperature between the cold water and that to 
which it is to be raised, for a dividend, then to the tempe- 
rature of the steam add 900 degrees, and from that sum 
take the required temperature of the water: this last re- 
mainder being made a divisor to the above dividend, the 
quotient will be the quantity of steam in the same terms 
as the water. 
Z 2 



270 STEAM ENGINE. 



What quantity of steam at 212° will raise 100 gallons 
of water at 60° up to 212°? 

212°_60° X 100 ,„ „ ^ r , - 

2r2- ^900°^212 == ^^ S'^^°"' ^^ ''^^'' ^^^"^"^ '"^^ 
steam. 

Now, steam at the temperature of 212° is 1800 times 
its bulk in water; or 1 cubic foot of steam, when its 
elasticity is equal to 30 inches of mercury, contains 1 
cubic inch of water. Therefore 17 gallons of water con- 
verted into steam, occupies a space of 4090| cubic feet, 
having a pressure of 15 lbs. on the square inch. 

In boiling by steam, using a jacket instead of blowing 
the steam into the water, about 10.5 square feet of sur- 
face are allowed for each horse capacity of boiler; that 
is, a 14 horse boiler will boil water in a pan set in a 
jacket, exposing a surface of 10.5 X 14 = 147 square feet. 

Horse Power. — Boulton and Watt suppose a horse able 
to raise 32,000 lbs. avoirdupois 1 foot high in a minute. 
Desaguliers makes it 27,500 lbs. 
Smeaton do. 22,916 do. 

It is common in calculating the power of engines, to 
suppose a horse to draw 200 lbs. at the rate of 2^ miles in 
an hour, or 220 feet per minute, with a continuance, draw- 
ing the w^eight over a pulley — now, 200 X 220 = 44000, 
i. e. 44000 lbs. at 1 foot peV minute, or 1 lb. at 44000 feet 
per minute. In the following table 32,000 is used. 

One horse power is equal to raise gallons or 

lbs, feet high per minute. 



STEAM ENGINE. 



271 



Feet High 


Ale 


Lbs. 


Feet Hish 


Ale 


Lbs. 


Per Minute. 


Callous. 


x\voiidupois. 


Per Minute. 


Gallons. 


Avoirdupois. 


I 


3123 


32000 


20 


156 


1600 


2 


1561i 


16000 


25 


125 


1280 


3 


1041 


10666 


30 


104 


1066 


4 


780 


8000 


35 


89 


914 


5 


624 


6400 


40 


78 


800 


6 


520 


5333 


45 


69 


711 


7 


446 


4571 


50 


62 


640 


8 


390 


4000 


55 


56 


582 


9 


347 


3555 


60 


52 


533 


10 


312 


3200 


65 


48 


492 


11 


284 


2909 


70 


44 


457 


12 


260 


2666 


75 


41 


426 


13 


240 


2461 


80 


39 


400 


14 


223 


2286 


85 


37 


376 


15 


208 


2133 


90 


34 


355 


16 


195 


2000 


95 


32 


337 


17 


183 


1885 


100 


31 


320 


18 


173 


1777 


110 


28 


291 


19 


164 


1684 


120 


26 


267 



Length of Stroke. — The stroke of an engine is equal 
to one revolution of the crank shaft, therefore double the 
length of the cylinder. When stating the lengtli of stroke, 
the length of cylinder is only given, that is, an engine 
with a 3 feet stroke, has its cylinder 3 feet long, besides 
an allowance for the piston. 

The following table shows the length of stroke, (or 
lengt!) of cylinder,) and the number of feet the piston 
travels in a minute, according to the number of strokes 
the engine makes when working at a maximum. 

When calculating the power of engines, the feet per 
minute are generally taken at 220. 



272 



STEAM EI^GINE. 



Length 
Stroke. 


Number 

of 
Strokes. 


Feet 

per 

Minute. 


Feet 2 


43 


172 


. . 3 


32 


192 


. . 4 


25 


200 


. . 5 


21 


210 


. . 6 


19 


228 


. . 7 


17 


238 


. . 8 


15 


240 


. . 9 


14 


250 



CYLI^'DEK. — When an engine in good order is perform- 
ing its regular work, the effective pressure may be taken 
at 8 Jbs. on each square inch of the surface of the piston. 

To calculate the power of an Engine. 

Rule 1. Multiply the area of cylinder by the effective 
pressure=say 8 lbs, the product is the weight the engine 
can raise. — Multiply this we'ght by the number of feet the 
piston travels in one minute, which will give the momen- 
tum, or weight, the engine can lift 1 foot high per minute; 
divide this momentum by a horse power, as previously 
stated, and the quolient will be the number of horse power 
the engine is equal to do. 

Rule 2. 25 inches of the area of cylinder is equal to 
one horse power, the velocity of the engine being con- 
stantly 220 feet per minute. 

EXAMPLE I. 

What is the power of an engine, tlie cylinder being 42 
inches diameter, and stroke 5 feet ? 
423 X .7854 X 10 X 210 
——^ =66.12 horse power. 

EXAMPLE 11. 

What size of cylinder will a 60 horse power engine re- 
quire, when the stroke is 6 feet ? 

44000 X 60 . , r .' . 

— ^oo V. .. =^^^^ inches area of cylinder. 
228 X 10 ^ 



STEAM ENGINE. 



273 



Note. To find the power to lift a weight at any velo- 
city, multiply the weight in lbs. by the velocity in feet, and 
divide by the horse power ; the quotient will be the num- 
ber of horse power required. 



Wlien tlje etfeci- 

ive piessiire on 

each inch of 

piston is 


The area equal to 

one horse power 

will be 


63 lbs. 


3.7 inches. 


48 — 


4.17 — 


43 — 


4.65 — 


38 — 


5.26 — 


33 — 


6.06 — 


28 — . 


7.14 — 


23 — 


8.7 — 


18 — 


11.11 — 


13 — 


15.46 — 


8 — 


25. — 



NozLEs.-T-The diameter of the valves of nozles ought 
to be fully one fifth of the diameter of cylinder. 

AiR-PuMP. — The solid contents of the air-pump is equal ' 
to the fourth of the solid contents of cylinder, or when the 
air-pump is half the length of the stroke of the engine, its 
area is equal to half the area of the cylinder. 

Condenser — is generally equal in capacity to the air- 
pump ; but when convenient, it ought to be more ; for 
when large, there is a greater space of vacuum, and the 
steam is sooner condensed. 

Cold Water Pump. — The capacity of the cold water 
pump depends on the temperature of the water. Many 
engines return their water, v/hich cannot be so cold as 
water newly drawn from a river, well, <fec. ; but when 
water is at the common temperature, each horse power 
requires nearly 7-^ gallons per minute. Taking this quan- 
tity as a standard, the size of the pump is easily found by 
the following rule, viz. — Multiply the number of horse 
power by '7|- gallons, and divide by the num.ber of strokes 
per minute ; this will give the quantity of water to be rais- 
ed each stroke of pump. Multiply this quantity by 231, 
(the number of cubic inches in a gallon,) and divide by 



274 STEAM ENGINE. 

the length of effective stroke of pump : the quotient will 
be the area. 

EXAMPLE. 

What diameter of pump is requisite for a 20 horse power 

steam engine having a 3 feet stroke, the effective stroke 

of pump to be fifteen inches ? 

20 X 71=150 , ^„„, 

^ -— =4.687o gallons the pump lifts each 

stroke. 

4.6875 X 231 



15 



= 72.1875 inches area of pump. 



Hot Water Pump. — The quantity of water raised at 
each stroke ought to be equal in bulk to the 900th part 
of the capacity of the cylinder. 

Proportions. — The length of stroke being 1, the length 
of beam to centre will be 2, the length of crank .5, and 
the length of connecting rod three. 

The following table shows the force which the connect- 
ing rod has to turn round the crank at different parts of 
the motion. 



Col. A. Decimal proportions of descent of 
the Piston, the whole descent being 
1. 



A 


B 


C 


D 


.0 


180° 


.0 


.0 


.05 


15U 


.46 


.128 


.10 


141 


.62 


.158 


.15 


13U 


.74 


.228 


.2 


123* 


.830 


.271 


.25 


117i 


.892 


.308 


.3 


1101 


.94 


.342 


.35 


104 


.976 


.377 


.4 


97* 


.986 


.41 


.45 


91f 


1. 


.441 


.5 


85* 


1. 


.473 


.55 


80 


.986 


.507 


.6 


75 


.956 


.538 


.65 


69 


.92 


.572 


.7 


62* 


.88 


.607 


.75 


57* 


.824 


.642 


.8 


49 


.746 


.68 


.85 


42 


.66 


.723 


.9 


34 


.546 


.776 


.95 


23* 


.390 


.84 


1.0 





.000 


1.0 



Col. B. Angle between the connecting Rod 
and Crank. 



Col. C. Effective length of the Lever upon 
which the connecting Rod acts, the 
whole Crank being 1. 



Col. D. Decimal proportions of half a revo- 
lution of the Fly- Wheel. 



Col. C also shows the force which is com- 
municated to the Fly-Wheel, ex- 
pressed in decimals, the force of 
the Piston being 1. 



STEAM ENGINE. 275 

Fly Wheel — Is used to regulate the motion of the en- 
gine, and to bring the crank past its centres. The rule 
for finding its weight is, — Multiply the number of horses' 
power of the engine by 2000, and divide by the square of 
the velocity of the circumference of the wheel per second : 
the quotient will be the weight in cwts. 

EXAMPLE. 

Required the weight of a fly-wheel proper for an engine 
of 20 horse power, 18 feet diameter, and making 22 revo- 
lutions per minute? 

18 feet diameter = 56 feet circumference, X 22 revo- 
lutions per minute = 1232 feet, motion per minute -:- 
60 = 201 feet motion per second ; then 20i2 ^ 4201 the 
divisor. 

20 horse power X 2000 = 40000 dividend. 

40000 

-— — = 90.4 cwt. weight of wheel. 

Parallel Motion. — The radius and parallel bars are 
of the SBme dimensions ; their length being generally 1.4 
of the length of the beam between the two glands, or one 
half of the distance between the fulcrum and gland. Both 
pairs of straps are the same length between the centres, 
and which is generally three inches less than the half of 
the length of stroke. 

Governor or Double Pendulum. — If the revolutions 
be the same, whatever be the length of the arms, the balls 
will revolve in the same plane, and the distance of that 
plane from the point of suspension, is equal to the length 
of a pendulum, the vibrations of which will be double the 
revolutions of the balls. For example : suppose the dis- 
tance between the point of suspension and plane of revo- 
lution be 36 inches, the vibrations that a pendulum of 36 

375 
inches will make per minute is,= -jTrr = 62 vibrations^ 

y/o6 

62 
and— -=31 revolutions per minute the balls ought to makeV 



276 WATER WriEEL. 

WATER WHEEL. 
Water. (^Hydrostatics) 

Hydrostatics is the science which treats of the pressure^ 
or weight, and equihbrium of water, and other fluids, es- 
pecially those that are non-elastic. 

Note 1. The pressure of water at any depth, is as its 
depth ; for the pressure is as the weight, and the weight 
is as the height. 

Note 2. The prisssure of water on a surface any how 
immersed in it, either perpendicular, horizontal, or ob- 
lique, is equal to the weight of a column of water, the 
base being equal to the surface pressed, and the altitude 
equal to the depth of the centre of gravity, of the surface 
pressed, below the top or surface of the fluid. 

PROBLEM I. 
In a vessel filled with water, the sides of which are up- 
right and parallel to each other, having the top of the 
same dimensions as the bottom, the pressure exerted 
against the bottom will be equal to the area of the bottom 
multiplied by the depth of water. 

EXA3irLE. 

A vessel 3 feet square and 7 feet deep is filled with wa- 
ter ; what pressure does the bottom support ? 
33 + 7 _L 1000 
■ -4 =39371 lbs. Avoirdupois. 

PROBLEM IL 
A side of any vessel sustains a pressure equal to the 
area of the side multiplied by half the depth, therefore the 
sides and bottom of a cubical vessel sustain a pressure 
equal to three times the weight of water in a vessel. 

EXAMPLE I. 

The gate of a sluice is 12 feet deep and 20 feet broad: 
what is the pressure of water against it? 
20X12x6X1000 
=90000=401 tons nearly. 

From Note 2d, The pressure exerted upon the side 
of a vessel, of whatever shape it may be, is as the area of 
the side and centre of gravity below the surface of waleri 



I 



WATER WHEEL. 277 

EXAMPLE II. 

Wlmt f)ressure will a board sustain, placed diagofially 

through a vessel, the side of which is 9 feet deep, and 

bottom 12 feet by 9 feet? 

\/ 122-f-9^=15 feet, the length of diagonal board. 

15X9X4^X1000' ^^„^^,, , 

=- =37969 lbs. nearly. 

16 ^ 

Though the diagonal board bisects the vessel, yet it 
sustains more than half of the pressure in the bottom, for 
the area of bottom is 12x9, and the half of the pressure 
is i of 60750=30375. 

The bottom of a conical or pyramidical vessel sustains 
a pressure equal to the area of the bottom and depth of 
water, consequently, the excess of pressure is three times 
the weight of water in the vessel. 

Water (.Hydraulics.) 

Hydrauhcs is that science which treats of fluids consi- 
dered as in motion; it therefore embraces the phenomena 
exhibited by water issuing from orifices in reservoirs, pro- 
jected obliquely, or perpendicularly, \n jets-d^eau, moving 
in pipes, canals, and rivers, oscillating in waves, or oppos- 
ing a resistance tc the progress of solid bodies. 

It would be needless here to go into the minutiee of 
hydraulics, particularly when the theory and practice do 
not agree. It is only the general laws, deduced from ex- 
periment, that can be safely employed in the various 
operations of hydraulic architecture. 

Mr. Banks, in his Treatise on Mills, after enumerating 
a number of experiments on the velocity of flowing water, 
by several philosophers, as well as his own, takes from 
thence the following simple rule, which is as near the truth 
as any that have been stated by other experimentalists. 

Rule. Measure the depth (of the vessel, &c.) in feet, 
extract the square root of that depth, and multiply it by 
5.4, which gives the velocity in feet per second; this 
multiplied by the area of the orifice in feet, gives the num- 
ber of cubic feet which flows out in one second. 

A A 



278 WATER WHEEL. 

EXAMPLE. 

Let a sluice be 10 feet below the surface of the water, 
its length 4 feet, and open 7 inches; required the quantity 
of water expended in one second? 

■v/ 10=3.162X5.4=17.0748 feet velocity. 

4X7 

— — = H feet X 17.0748=39.84 cubic feet of water 

per second. 

If the area of the orifice is great compared with the 
head, take the medium depth, and two thirds of the veloci- 
ty from that depth, for the velocity. 

EXAMPLE. 

Given the perpendicular depth of the orifice 2 feet, its 
horizontal length 4 feet, and its top 1 foot below the sur- 
face of water. To find the quantity discharged in one 
second: 

The medium depth is = 1.5 x 6.4 = 8.10 f of 8.10 = 
5.40, and 5.40 X 8 = 43.20 cubic feet.* 

'J'he quantity of water discharged through slits, or notch- 
es, cut in the side of a vessel or dam, and open at the top, 
will be found by multiplying the velocity at the bottom by 
the depth, and taking f of the product for the area; which 
again multiplied by the breadth of the slit or notch, gives 
the quantity of cubic feet discharged in a given time. 

EXAMPLE. 

Let the depth be 5 inches, and the breadth 6 inches; 
required the quantity run out in 46 seconds? 
The depth is .4166 of a foot. 
The breadth is .5 of a foot. 

^/.4166=.6445X5.4X| = 2.3238X.4166 = .96825 
X.5=. 48412 feet per second. 

Then .48412x46=222.69 cubic feet in 46 seconds. 

There are two kinds of water wheels, Undershot and 
Overshot. Undershot, when the water strikes the wheel 
at, or below the centre. Overshot when the water fills 
upon the wheel above the centre. 

* The square root of the depth is not taken in this example, but 
when the depth is considerable, it ought to be taken. 



WATER WHEEL. 279 

The effect produced by an undershot wheel, is from the 
impetus of the water. The effect produced by an overshot 
wheel, is from the gravity or weight of the water. 

Of an undershot wheel, the power is to the effect as 3: 1 . 
Of an overshot wheel, the power is to the effect as 3: 2 — 
which is double the effect of an undershot wheel. 

The velocity at a maximum is=:3 feet in one second. 

Since the effect of the overshot is double that of the 
undershot, it follows that the higher the wheel is in pro- 
portion to the whole descent, the greater will be the effect. 

The maximum load for an overshot wheel is that which 
reduces the circumference of the wheel to its proper 
velocity, = 3 feet in one second; and this will be known, 
by dividing the effect it ought to produce in a given time, 
by the space intended to be described by the circumference 
of the wheel in the same time; the quotient will be the re- 
sistance overcome at the circumference of the wheel, and 
is equal to the load required, the friction and resistance 
of the machinery included. 

The following is aji extract from Banks on Mills. 

The effect produced by a given stream in falling through 
a given space, if compared with a weight, will be directly 
as that space; but if we measure it by the velocity com- 
municated to the wheel, it will be as the square root of 
the space descended through, agreeably to the laws of 
falling bodies. 

Experiment 1. A given stream is applied to a wheel at 
the centre; the revolutions per minute are 38.5. 

Ex. 2. The same stream applied at the top, turns the 
same wheel 57 times in a minute. 

If in the first experiment the fall is called 1, in the se- 
cond it will be 2: then the ^/l : s/2 :: 38.5 : 54.4, which 
are in the same ratio as the square roots of the spaces 
fallen through, and near the observed velocity. 

In the following experiments a fly is connected with the 
water wheel. 

Ex. 3. The water is applied at the centre, the wheel 
revolves 13.03 times in one minute. 



280 



WATER WHEEL. 



Ex. 4. The water is applied at the vertex of the wheel, 
and it revolves 18.2 times per minute. 

As 13.03 : 18.2 :: -y/l : y,/2 nearly. 

From the above we infer, that the circumferences of 
wheels of different sizes may move with velocities which 
are as the square roots of their diameters without disad- 
vantage, compared one with another, the water in all be- 
ing applied at the top of the wheel, for the velocity of 
falling water at the bottom or end of the fall is as the 
time, or as the square root of the space fallen through; for 
example, let the fall be 4 feet, then, As-v/16 ; 1" :: >/4 : ^", 
the time of falling through 4 feet: — Again, let the fall be 
9 feet, then, ^]6: 1"::\/9:|", and so for any other 
space, as in the following table, where it appears that 
water will fall through one foot in a quarter of a second, 
through 4 feet in half a second, through 9 feet in 3 quar- 
ters of a second, and through 16 feet in one second. And 
if a wheel 4 feet in diameter moved as fast as the water, 
it could not revolve in less than 1.5 second, neither could 
a wheel of 16 feet diameter revolve in less than three se- 
conds; but though it is impossible for a wheel to move as 
fast as the stream which turns it; yet, if their velocities 
bear the same ratio to the time of the fall through their 
diameters, the wheel 16 feet in diameter may move twice 
as fast as the wheel 4 feet in diameter. 



Height 


Time of 


Height 


Time of 


of the fall 


falling in 


of the fall 


falling in 


in feet. 


seconds. 


in feet. 


seconds. 


1 


.25 


14 


.935 


2 


.352 


16 


1. 


3 


.432 


20 


1.117 


4 


.5 


24 


1.22 


5 


.557 


25 


1.25 


6 


.612 


30 


1.37 


7 


.666 


36 


1.5 


8 


.706 


40 


1.58 


9 


.75 


45 


1.67 


10 


.79 


50 


1.76 


12 


.864 







WATER WHEEL. 281 

Power akd effect. — The power water has to produce 
mechanical effect, is as the quantity and fall of perpen- 
dicular height. — The mechanical effect of a wheel is as 
the quantity of water in the buckets and the velocity. 

The power is to the effect as 3 : 2, that is, suppose the 
power to be 9000, the effect will be 
__9000 X 2_ 1 8000_ 
3 ■" "3 "~ 

Height of the wheel. — The higher the wheel is in 
proportion to the fall, the greater will be the effect, because 
it depends less upon the impulse, and more upon the gra- 
vity of the water; however, the head should be such, that 
the water will have a greater velocity than the circumfe- 
rence of the wheel; and the velocity that the circumfe- 
rence of the wheel ought to have, being known, the head 
required to give the water its proper velocity, can easily 
be known from the rules of Hydrostatics. 

Velocity of the wheel. — Banks, in the foregoing 
quotation, says, That the circumferences of overshot wheels 
of different sizes may move with velocities as the square 
roots of their diameters, without disadvantage. Smeaton 
says. Experience confirms that the velocity of 3 feet per 
second is applicable to the highest overshot wheels, as 
well as the lowest; though high wheels may deviate fur- 
ther from this rule, before they will lose their power, by a 
given aliquot part of the whole, than low ones can be ad- 
mitted to do; for a 24 feet wheel may move at the rate of 
6 feet per second, without losing any considerable part of 
its power. 

It is evident that the velocities of wheels will be in pro- 
portion to the quantity of water and the resistance to be 
overcome: — if the water flows slowly upon the wheel, 
more time is required to fill the buckets than if the water 
flowed rapidly; and whether Smeaton or Banks is taken 
as a data, the mill-wright can easily calculate the size of 
his wheel, when the velocity and quantity of water in a 
given time is known. 

EXAMPLE I. 

What power is a stream of water equal to, of the follow- 

A a2 



282 WATER WHEEL. 

ing dimensions, viz. 12 inches deep, 22 inches broad; ve- 
locity, 70 feet in llf seconds, and fall, 60 feet? — Also, 
what size of a wheel could be applied to this fall? 

12x22 

-— — — = 1.83 square feet: — area of stream. 
144 

ll|r" : 70 :: 60": 357.5 lineal feet per min. — velocity. 

357.5 X 1.83=654.225 cubic feet per minute. 

654.225X62.5=40889.0625 avoir, lbs. per minute. 

40889;0625X 60=2453343.7500 momentum at a fall of 

[60 feet. 

2453343.7500 ^^ ^ ^ 

44000 =^^-^ ^""''^ P""'"^'- 

3:2:: 55.7 : 37.13 effective power. 

The diameter of a wheel applicable to this fall, will be 

58 feet, allowing one foot below for the water to escape, 

and one foot above for its free admission. 

58x3.1416 = 182.2128 circumference of wheel. 

60x6=360 feet per minute,=velocity of wheel. 

654.225 „ . , ,, , 

— — — — = 1.8 sectional area ot buckets. 
obO 

The bucket must only be half full, therefore 1.8x2= 
3.6 will be the area. 

To give sufficient room for the water to fill the buckets, 
the wheel requires to be 4 feet broad. 

Now, -^=.9, say 1 foot depth of shrouding. 

360 
-TT-T-— -— = 1.9 revolutions per min. the wheel will make. 
182.2128 

Power of water . . =55.7 h. p. ") 

Effective power of do. =:37.13 h. p. J 

Dimensions C Diameter . . =58 feet. [ Ans» 

of ? Breadth . . = 4 feet. | 

Wheel. (Depth ofshrouding= 1 foot. J 

EXAMPLE II. 

What is the power of a water wheel, 16 feet diameter, 
12 feet wide, and shrouding 15 inches deep? 



WATER WHEEL. 283 

16x3.1416=50.2656 circumference of wheel. 
12 X 11 = 15 square feet, sectional area of buckets. 
60x4=240 lineal feet per minute, =velocity. 
240x15=3600 cubic feet water, when buckets are full; 

when half full, 1800 cubic feet. 
1800x62.5=112500 avoir, lbs. of water per minute. 
112500x16 = 1800000 momentum, falling 16 feet. 

1200000 

3:2:: 1800000 :-——-— =27 horse power. 

44000 

Buckets. — The number of buckets to a wheel should be as few 
as possible, to retain the greatest quantity of water ; and their 
mouths only such a width as to admit the requisite quantity of 
water, and at the same time sufficient room to allow the air to 
escape. 

The Communication of Power. — There are no prime movers of 
machinery, from which power is taken in a greater variety of 
forms than the water wheel ; and among such a number there 
cannot fail to be many bad applications. 

Suffice it here to mention one of the worst, and most generally 
adopted. For driving a cotton mill in this neighbourhood, there 
is a water wheel about 12 feet broad, and 20 feet diameter; there 
is a division in the middle of the buckets, upon which the seg- 
ments are bolted round the wheel, and the power is taken from 
the vertex; from this erroneous application, a great part of the 
power is lost; for the weight of water upon the wheel presses 
against the axle in proportion to the resistance it has to overcome, 
and if the axle was not a very large mass of wood, with very strong 
iron jovrnals, it could not stand the great strain which is upon it. 

The most advantageous part of the wheel, from which the power 
can be taken, is that point in the circle of gyration horizontal 
to the centre of the axle; because, taking the power from this 
part, the whole weight of water in the buckets acts upon the teeth 
of the wheels; and the axle of the water wheel suffers no strain. 

The proper connection of machinery to water wheels is of the 
first importance, and mismanagement in this particular point is 
often the cause of the journals and axles giving way, besides a 
considerable loss of power. 

EXAMPLE. 

Required the radius of the circle of gyration in a water 
wheel, 30 feet diameter; the weight of the arms being 12 
tons, shrouding 20 tons, and water 15 tons. 



284 PUMPS. 

30 feet diameter, radius=15 feet. 
S 20X162=4500X2=9000^ The opposite side of the 
A22X15_^_ 900X2=1800J ^^^^^^^^ ''^''^ "^"^^ ^« 

W 15x162=3375 =3375 
2X(20+12)=64 nr75- 

W 15 itq = 179, the square root of 

79 
which is 13y\ feet, the radius of the circle of gyration. 

PUMPS. 

There are two kinds of Pumps, Lifting and Forcing. 
The Lifting, or Common Pumps, are appHed to wells, &c. 
where the depth does not exceed 32 feet; for beyond this 
depth they cannot act, because the height that water is 
forced up into a vacuum, by the pressure of the atmos- 
phere, is about 34 feet. 

The Force Pumps are those that are used on all other 
occasions, and can raise water to any required height. — 
Bramah's celebrated pump is one of this description, and 
shows the amazing power that can be produced by such 
application, and which arises from the fluid and non-com- 
pressible qualities of water. 

The power required to raise water any height is equal 
to the quantity of water discharged in a given time, and 
the perpendicular height. 



Required the power necessary to discharge 175 ale gal- 
lons of water per minute, from a pipe 252 feet high? 

One ale gallon of water weighs 10^: lbs. avoir, nearly. 
175X10^=1799X252=453348 ^^ „ ^ 

-^^—=10.3 horse power. 

The following is a very simple rule, and easily kept in 
remembrance. 

Square the diameter of the pipe in inches, and the pro- 
duct will be the number of lbs. of water avoirdupois con- 
tained in every yard length of the pipe. If the last figure 



PUMPS. 285 

of the product be cut off, or considered a decimal, the re- 
maining figures will give the number of ale gallons in 
each yard of pipe; and if the product contains only one 
figure, it will be tenths of an ale gallon. The number of 
ale gallons multiplied by 282, gives the cubic inches in 
each yard of pipe; and the contents of a pipe may be 
found by Proportion. 

EXAMPLE. 

What quantity of water will be discharged from a pipe 

5 inches diameter, 252 feet perpendicular height, the water 
flowing at the rate of 210 feet per minute? 

210 

5"^ X-^ = 175 ale gallons per minute. 

252 
5^ X-^=2100 lbs. water in a pipe. 

2100 X 210 ^^^ ^ 

: ^^^ — =10 horse power required to pump that 

44000 r r 

quantity of water. 

The following table gives the contents of a pipe one • 
inch in diameter, in weight and measure; which serves as 
a standard for pipes of other diameters, their contents 
being found by the following rule. 

Multiply the numbers in the following table against any 
height, by the square of the diameter of the pipe, and the 
product will be the number of cubic inches, avoirdupois 
ounces, and wine gallons of water, that the given pipe 
will contain. 

EXAMPLE. 

How many wine gallons of water is contained in a pipe 

6 inches diameter, and 60 feet long? 

2.4480X36=88.1280 wine gallons. 
In a wine gallon there are 231 cubic inches. 



286 



OxVE INCH DIAMETER. | 


Feet 


Uuaiitity in 


Weight in 


Gallons 


High. 


Cubic Inches. 


Avoir. Oz. 


Wine Measure. 


1 


9.42 


5.46 


.0407 


2 


18.85 


10.92 


.0816 


3 


28.27 


16.38 


.1224 


4 


37.70 


21.85 


.1632 


5 


47.12 


27.31 


.2040 


6 


56.55 


32.77 


.2448 


7 


65.97 


38.23 


,2423 


8 


75.40 


43.69 


.3264 


9 


84.82 


49.16 


.3671 


10 


94.25 


54.62 


.4080 


20 


188.49 


109.24 


.8160 


30 


282.74 


163.86 


1.2240 


40 


376.99 


218.47 


1.6300 


50 


471.24 


273.09 


2.0400 


60 


565.49 


327.71 


2.4480 


70 


659.73 


382.33 


2.8560 


80 


753.98 


436.95 


3.2640 


90 


848.23 


491.57 


3.6700 


100 


942.48 


546.19 


4.0800 


200 


1884.96 


1092.38 


8.1600 



The resistance arising from the friction of water flow- 
ing through pipes, &c. is directly as the velocity of the 
water, and inversely as the circumference of the pipe. 

The data given is a medium, and which is |th of the 
whole resistance; this is the standard generally adopted, 
being considered as mosfcorrect. 

EXAMPLE I. 

What is the power requisite to overcome the resistance 
and friction of a column of water 4 inches diameter, 100 
feet high, and flowing at the velocity of 300 feet per mi- 
nute? 

546.19 X42 

=546.19, say 546.2 

16 



546.2x30 

44000 
required to overcome the resistance occasioned by the 



= 3.7, ith of which is .7, therefore the power 



PUMPS. 287 

weight and friction of the water will be 3. 7 + . 7=4. 4 H. 
P., say 4.5 horse power. 

EXAMPLE II. 

There is a cistern 20 feet square, and 10 feet deep, 

placed on the top of a tower 60 feet high, what power is 

requisite to fill this cistern in 30 minutes, and what will be 

the diameter of the pump, when the length of stroke is 2 

feet, and making 40 per minute? 

20X20X 10=4000 cubic contents of cistern. 

4000 

-——=133.3 cubic feet of water per minute. 

133.3x1000 ^^^^ ^^ ,, . J . 

=8331.2o lbs. avoirdupois per minute. 

16 

8331.25X60 , , „^ , i . . r u- i, 

= lI.o6 horse power, l-oili oi which 

44000 ^ 

is=2. 27+11.1 1 = 13.63 horse power required. 

133 3 

^ = 1.7x144=244.80 

2X40=:80 ^-'^^^^ _^ =.311.7, now 

.7854 



y/31 1.7 = 17.6 inches diameter of pump required. 

Founders generally prove the pipes they cast to stand a certain 
pressure, which is calculated by the weight of a perpendicular 
column of water, the area being equal to the area of the pipe, 
and the height equal to any given height. 

To ascertain the exact pressure of water to which a pipe is 
subjected, a safety valve is used, generally of 1 inch diameter 
and loaded with a weight equal to the pressure required : for ex- 
ample, a pipe requires to stand a pressure of 300 feet, what weight 
will be required to load the safety valve one inch diameter? 
Feet. Inches. Ounces. 

300 X 12= 3600 X .7854 = 2827.4400 X 1000 ,^„^, 

VTOR =1636i ,^ 

-Jg— =102 
lbs. 4i oz. weight required. 

Each of the weights for the safety valves of these hydrostatic 

proving machines are generally made equal to a pressure of a co- 

luran of water 50 feet high, the area being the area of the valve. 

50 feet of pressure on a valve 1 inch diam. = 17.06 lbs. 

50 do. do. do. H do. = 26.65 do. 

50 do. do. do. 1^ do. =38.38 do. 

50 do. do. do. 2 do. = 68.24 do. 

In pumping, there is always a deficiency owing to the escape 



of water through the valves ; lo account for this loss, there is an 
allowance of 3 inches for each stroke of piston rod : for example, 
a three feet stroke may be calculated at 2 feet 9 inches. 

There is a town, the inhabitants of which amount to 12000, 
and it is proposed to supply it with water, from a river running 
through the low grounds 250 perpendicular feet below the best 
situation from the reservoir. 

It is required to know the power of an engine capable of lifting 
a sufficient quantity of water, the daily supply being calculated 
at 10 ale gallons to each individual : also what size of pump and 
pipes are requisite for such ? 

12000 X 10 =120000 gallons per day. 

Engine is to work 12 hours, — — — = 10000 gallons per hour. 

10000 ,^,, „ 

^^ 166.6 gallons per minute. 

The pump to have an effective stroke of 3| feet, and making 30 
strokes per minute. 

' =5.5533 gallons each stroke. 

282 X 5.6 = 1579.2 cubic inches of water each stroke. 
1579.2 „, , . , . 

o/. .«• , —r^- — = 35.1 inches, area of pump. 
3feet9inches=45in. ^ ^ 

35 ]^ 

— g^=44.7, therefore V 44.1 = 6.7 diameter of pump. 

The pipes will require to be at least the diameter of the pump; 
if they are a little more, the water will not require to flow so 
quickly through them, and thereby cause less friction. 
The power of the engine will be 

166.6 gall. X lOi lb. x 250 feet = 426925 momentum. 

————=9.7, add l-5th = 11.64 horse power. 
44000 

iis=>^=' ''■'' '■°-'^'-"- 

426925 ,__ T„- , „ ,. 

— — —=15.5, = lo.6 do. Desaguhers. 

^22916"" ^^•^' ^ ^^'^^ ^°' ^'^^^^°^' 



in My '08 



I 



